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a: Khi x=-2 thì \(y=-3\cdot\left(-2\right)^2=-12\)
Khi x=-1 thì \(y=-3\cdot\left(-1\right)^2=-3\)
Khi x=-1/3 thì \(y=-3\cdot\dfrac{1}{9}=-\dfrac{1}{3}\)
Khi x=0 thì y=0
Khi x=1/3 thì \(y=-3\cdot\dfrac{1}{9}=-\dfrac{1}{3}\)
Khi x=1 thì y=-3
Khi x=2 thì y=-12
b: 
\(\dfrac{1}{x^2+2}-\dfrac{1}{xy+2}+\dfrac{1}{y^2+2}-\dfrac{1}{xy+2}=0\)
\(\Leftrightarrow\dfrac{xy-x^2}{\left(x^2+2\right)\left(xy+2\right)}+\dfrac{xy-y^2}{\left(y^2+2\right)\left(xy+2\right)}=0\)
\(\Leftrightarrow\dfrac{x-y}{xy+2}\left(\dfrac{y}{y^2+2}-\dfrac{x}{x^2+2}\right)=0\)
\(\Leftrightarrow\left(\dfrac{x-y}{xy+2}\right)\left(\dfrac{x^2y+2y-xy^2-2x}{\left(x^2+2\right)\left(y^2+2\right)}\right)=0\)
\(\Leftrightarrow\dfrac{\left(x-y\right)^2\left(xy-2\right)}{\left(xy+2\right)\left(x^2+2\right)\left(y^2+2\right)}=0\)
\(\Leftrightarrow xy=2\) (do x;y phân biệt)
\(\Rightarrow P=\dfrac{2}{xy+2}+\dfrac{2}{xy+2}=\dfrac{4}{xy+2}=\dfrac{4}{2+2}=1\)
a: \(A=\dfrac{2\sqrt{x}+6+\sqrt{x}-3}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{3\left(\sqrt{x}+1\right)}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{3}{\sqrt{x}+3}\)
b: \(\sqrt{x}+3>=3\)
=>A<=1
Dấu = xảy ra khi x=0
c: \(P=A:\left(B-1\right)=\dfrac{3}{\sqrt{x}+3}:\dfrac{2\sqrt{x}+1-\sqrt{x}-3}{\sqrt{x}+3}=\dfrac{3}{\sqrt{x}-2}\)
Để P nguyên thì căn x-2\(\in\left\{1;-1;3;-3\right\}\)
=>\(x\in\left\{1;25\right\}\)
Câu 1:
a)
| \(y=f\left(x\right)=2x^2\) | -5 | -3 | 0 | 3 | 5 |
| f(x) | 50 | 18 | 0 | 18 | 50 |
b) Ta có: f(x)=8
\(\Leftrightarrow2x^2=8\)
\(\Leftrightarrow x^2=4\)
hay \(x\in\left\{2;-2\right\}\)
Vậy: Để f(x)=8 thì \(x\in\left\{2;-2\right\}\)
Ta có: \(f\left(x\right)=6-4\sqrt{2}\)
\(\Leftrightarrow2x^2=6-4\sqrt{2}\)
\(\Leftrightarrow x^2=3-2\sqrt{2}\)
\(\Leftrightarrow x=\sqrt{3-2\sqrt{2}}\)
hay \(x=\sqrt{2}-1\)
Vậy: Để \(f\left(x\right)=6-4\sqrt{2}\) thì \(x=\sqrt{2}-1\)
a) ĐKXĐ: a\(\ge\)0, a\(\ne\)1
A=(\(\dfrac{\sqrt{a}+2}{\left(\sqrt{a}+1\right)^2}-\dfrac{\sqrt{a}-2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\)).\(\dfrac{\sqrt{a}+1}{\sqrt{a}}\)
A=\(\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)}\).\(\dfrac{\sqrt{a}+1}{\sqrt{a}}\)
A=\(\dfrac{2\sqrt{a}}{\left(\sqrt{a}+1\right)\left(a-1\right)}.\dfrac{\sqrt{a}+1}{\sqrt{a}}\)=\(\dfrac{2}{a-1}\)
b) Để A\(\in\)Z\(\Rightarrow\)x-1\(\in\) Ư(2)=\(\left\{-1,1,-2,2\right\}\)
| x-1 | -2 | -1 | 1 | 2 |
| x | -1 | 0 | 2 | 3 |
vì x\(\ge\)0,x\(\ne\)1 nên x\(\in\)\(\left\{-1,0,2,3\right\}\)
Hằng đẳng thức:
\(\left(x-y-z\right)^2=x^2+y^2+z^2+2\left(yz-xy-zx\right)=x^2+y^2+z^2-2\left(xy+xz-yz\right)\)
\(\Rightarrow x^2+y^2+z^2=\left(x-y-z\right)^2+2\left(xy+xz-yz\right)\)
Giờ thay \(x=\dfrac{1}{a}\) ; \(y=\dfrac{1}{b}\); \(z=\dfrac{1}{c}\) là ra cái người ta làm
\(P=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...\)
\(\Rightarrow2P=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...\)
\(\Rightarrow2P=1+\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...\right)\)
\(\Rightarrow2P=1+P\)
\(\Rightarrow P=1\).
Ta có : \(P=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...\)
<=> 2P = \(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+....=1+P\)
<=> P = 1
=> Bạn A nói gần đúng ; P = 1