Bài tập theo tuần Toán 9 - Tuần 7

$\sqrt[3]{x-2}=3\iff x-2=3^3\iff x=29.$ Vậy x = 29.

Ta có

$\begin{array}{l}{x}^3+y^3+z^3={\left(x+y\right)}^3+z^3-3xy(x+y)\\ =\left(x+y+z\right)\left[{\left(x+y\right)}^2-\left(x+y\right)z+z^2\right]-3xy\left(-z\right)(dox+y+z=0)\\ =0+3xyz=3xyz\end{array}$

$\begin{array}{l}A=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\left(A>0\right)\\ \Rightarrow A\sqrt{2}=\sqrt{2}\left(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\right)\\ =\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}\\ =\sqrt{{\left(\sqrt{7}\right)}^2+2.\sqrt{7}.1+1^2}-\sqrt{{\left(\sqrt{7}\right)}^2-2.\sqrt{7}.1+1^2}\\ =\sqrt{{\left(\sqrt{7}+1\right)}^2}-\sqrt{{\left(\sqrt{7}-1\right)}^2}\\ =\sqrt{7}+1-\left|\sqrt{7}-1\right|=\sqrt{7}+1-\left(\sqrt{7}-1\right)=2\Rightarrow A=\sqrt{2}(doA>0)\end{array}$

$\begin{array}{l}a)G=\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{1}{1-x}-\frac{2\sqrt{x}}{x-1}\right).\left(\sqrt{x}+1\right)\\ =\frac{\sqrt{x}\left(\sqrt{x}-1\right)+1-2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\left(\sqrt{x}+1\right)=\frac{x-\sqrt{x}+1-2\sqrt{x}}{\sqrt{x}-1}=\frac{x-3\sqrt{x}+1}{\sqrt{x}-1}\end{array}$

$\begin{array}{l}b)G=2\iff \frac{x-3\sqrt{x}+1}{\sqrt{x}-1}=2\iff x-3\sqrt{x}+1=2\sqrt{x}-2\\ \iff x-5\sqrt{x}+3=0\iff \left[\begin{array}{l}x=\frac{5+\sqrt{13}}{2}\\ x=\frac{5-\sqrt{13}}{2}\end{array}\right.\end{array}$

$\begin{array}{l} A=\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\\ =\sqrt[3]{{\left(\sqrt{2}\right)}^3+3{\left(\sqrt{2}\right)}^2.1+3\sqrt{2}{.1}^2+1}-\sqrt[3]{{\left(\sqrt{2}\right)}^3-3.{\left(\sqrt{2}\right)}^2.1+3.\sqrt{2}{.1}^2-1}\\ =\sqrt[3]{{\left(\sqrt{2}+1\right)}^3}-\sqrt[3]{{\left(\sqrt{2}-1\right)}^3}=\sqrt{2}+1-\sqrt{2}+1=2\end{array}$

$\sqrt{x-1}+\sqrt{x+2}=\sqrt{2x-3}$

$\begin{array}{l}a)A=\left(\frac{\sqrt{x}+2}{2-\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x+2}}-\frac{4x+2\sqrt{x}-4}{x-4}\right):\left(\frac{-2}{2-\sqrt{x}}+\frac{3+\sqrt{x}}{2\sqrt{x}-x}\right)\left(\begin{array}{l}x\ge 0\\ x\ne 4\end{array}\right)\\ =\frac{-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)+\sqrt{x}.\left(\sqrt{x}-2\right)-4x-2\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}:\frac{-2\sqrt{x}+3+\sqrt{x}}{\sqrt{x}.\left(2-\sqrt{x}\right)}\\ =\frac{-x-4\sqrt{x}-4+x-2\sqrt{x}-4x-2\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}.\frac{-\sqrt{x}.\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\\ =\frac{\left(-4x-8\sqrt{x}\right).\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\frac{-4\sqrt{x}.\left(\sqrt{x}+2\right).\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\frac{-4x}{\sqrt{x}-3}\end{array}$

$\begin{array}{l}b)A=\sqrt{x}+3\\ \iff \frac{-4x}{\sqrt{x}-3}=\sqrt{x}+3\iff x-9=-4x\iff -3x=-9\iff x=3\left(tm\right)\end{array}$

Vậy x = 3 thì $A=\sqrt{x}+3$

$\begin{array}{l}a)P=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}}{x-2\sqrt{x}+1}\left(\begin{array}{l}x>0\\ x\ne 1\end{array}\right)\\ =\frac{1+\sqrt{x}}{\sqrt{x}.\left(\sqrt{x}-1\right)}.\frac{{\left(\sqrt{x}-1\right)}^2}{\sqrt{x}}=\frac{\left(1+\sqrt{x}\right)\left(\sqrt{x}-1\right)}{x}=\frac{x-1}{x}\\ b)P>\frac{1}{2}\iff \frac{x-1}{x}>\frac{1}{2}\iff \frac{x-1}{x}-\frac{1}{2}>0\\ \iff \frac{2x-2-x}{2x}>0\iff x-2>0\iff x>2\left(tm\right)\end{array}$

Vậy x > 2 khi $P>\frac{1}{2}$

$\begin{array}{l}a)A=\left(1-\frac{2\sqrt{a}}{a+1}\right):\left(\frac{1}{\sqrt{a}+1}-\frac{2\sqrt{a}}{a\sqrt{a}+\sqrt{a}+a+1}\right)\left(\begin{array}{l}a\ge 0\\ a\ne 1\end{array}\right)\\ =\frac{a+1-2\sqrt{a}}{a+1}:\frac{a+1-2\sqrt{a}}{\left(\sqrt{a}+1\right)\left(a+1\right)}\\ =\frac{a+1-2\sqrt{a}}{a+1}.\frac{\left(a+1\right)\left(\sqrt{a}+1\right)}{a+1-2\sqrt{a}}=\sqrt{a}+1\\ b)a=2011-2\sqrt{2010}\Rightarrow \sqrt{a}=\sqrt{2010-2\sqrt{2010}.1+1}\\ =\sqrt{{\left(\sqrt{2010}-1\right)}^2}=\sqrt{2010}-1\\ \Rightarrow A=\sqrt{a}+1=\sqrt{2010}-1+1=\sqrt{2010}\end{array}$

Vậy $A=\sqrt{2010}$ khi $a=2011-2\sqrt{2010}$

$\begin{array}{l}a)B=\frac{2\left(x+4\right)}{x-3\sqrt{x}-4}+\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{8}{\sqrt{x}-4}\\ =\frac{2x+8}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}+\frac{\sqrt{x}.\left(\sqrt{x}-4\right)-8\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\end{array}$

$\begin{array}{l}=\frac{2x+8+x-4\sqrt{x}-8\sqrt{x}-8}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}=\frac{3x-12\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}\\ =\frac{3\sqrt{x}\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}=\frac{3\sqrt{x}}{\sqrt{x}+1}\end{array}$

$b)B=\frac{3\sqrt{x}}{\sqrt{x}+1}=\frac{3\sqrt{x}+3-3}{\sqrt{x}+1}=3-\frac{3}{\sqrt{x}+1}$

Để $B\in \mathbb{Z}\Rightarrow 3⋮\left(\sqrt{x}+1\right)\Rightarrow \sqrt{x}+1\in U\left(3\right)=\left\{1;3\right\}$ do $\sqrt{x}+1>0$

$\Rightarrow \sqrt{x}\in \left\{0;2\right\}\Rightarrow x\in \left\{0;4\right\}$

a)