a: Khi x=2 và y=-3 thì \(x^2+2y=2^2+2\cdot\left(-3\right)=4-6=-2\)

b: \(A=x^2+2xy+y^2=\left(x+y\right)^2\)

Khi x=4 và y=6 thì \(A=\left(4+6\right)^2=10^2=100\)

c: \(P=x^2-4xy+4y^2=\left(x-2y\right)^2\)

Khi x=1 và y=1/2 thì \(P=\left(1-2\cdot\dfrac{1}{2}\right)^2=\left(1-1\right)^2=0\)

a)\(\left(\frac{1-x^3+1-x-x}{1-x}\right):\frac{-\left(x-1\right)\left(x+1\right)}{\left(x+1\right)\left(x-1\right)^2}=\left(\frac{-x^3-2x+2}{1-x}\right)\cdot\left(1-x\right)=-x^3-2x+2\)

b) \(-\left(-1\frac{2}{3}\right)-2\cdot\left(-1\frac{2}{3}\right)+2=\frac{5}{3}+\frac{10}{3}+2=7\)

Tham khảo thanh này để soạn đề chính xác hơn nha :vvv

a) Ta có: \(M=\left(\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\right)\cdot\dfrac{x+3\sqrt{x}}{7-\sqrt{x}}\)

\(=\left(\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{7-\sqrt{x}}\)

\(=\dfrac{x-9-\left(x-2\sqrt{x}+\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{7-\sqrt{x}}\)

\(=\dfrac{x-9-x+\sqrt{x}+2}{\left(\sqrt{x}-2\right)}\cdot\dfrac{1}{-\left(\sqrt{x}-7\right)}\)

\(=\dfrac{\sqrt{x}-7}{\sqrt{x}-2}\cdot\dfrac{-1}{\sqrt{x}-7}\)

\(=\dfrac{-1}{\sqrt{x}-2}\)(1)

b) Ta có: \(x^2-4x=0\)

\(\Leftrightarrow x\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=4\left(loại\right)\end{matrix}\right.\)

Thay x=0 vào biểu thức (1), ta được:

\(M=\dfrac{-1}{\sqrt{0}-2}=\dfrac{-1}{-2}=\dfrac{1}{2}\)

Vậy: Khi \(x^2-4x=0\) thì \(M=\dfrac{1}{2}\)

Ta có: \(\dfrac{x^2-2x-3}{x^2+2x+1}=\dfrac{x^2+x-3x-3}{\left(x+1\right)^2}=\dfrac{x\left(x+1\right)-3\left(x+1\right)}{\left(x+1\right)^2}\)

\(=\dfrac{\left(x+1\right)\left(x-3\right)}{\left(x+1\right)^2}=\dfrac{x-3}{x+1}\left(dk:x\ne-1\right)\) (1)

Với \(x\ne-1\), ta có:

\(3x-1=0\Rightarrow3x=1\) \(\Rightarrow x=\dfrac{1}{3}\left(tm\right)\)

Thay \(x=\dfrac{1}{3}\) vào (1), ta được:

\(\dfrac{\dfrac{1}{3}-3}{\dfrac{1}{3}+1}=\left(\dfrac{1}{3}-3\right):\left(\dfrac{1}{3}+1\right)\)

\(=-\dfrac{8}{3}:\dfrac{4}{3}=-\dfrac{8}{3}\cdot\dfrac{3}{4}=-2\)

Vậy: ...

a; \(A=\left(\frac{x}{x^2-4}+\frac{2}{2-x}-\frac{1}{x+2}\right)\left(x+2\right)\)

\(=\frac{x-2\left(x+2\right)-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\cdot\left(x+2\right)=\frac{x-2x-4-x+2}{\left(x-2\right)}=\frac{-2x-2}{x-2}\)

Vì x=-2 không thỏa mãn ĐKXĐ
nên A không có giá trị khi x=-2

Thay x=0,5 vào A, ta được:

\(A=\frac{-2\cdot0,5-2}{0,5-2}=\frac{-3}{-1,5}=2\)

b: A là số nguyên

=>-2x-2⋮x-2

=>-2x+4-6⋮x-2

=>-6⋮x-2

=>x-2∈{1;-1;2;-2;3;-3;6;-6}

=>x∈{3;1;4;0;5;-1;8;-4}

a: \(A=\dfrac{x^2-8x+16-x^2+16}{\left(x-4\right)\left(x+4\right)}\cdot\dfrac{x}{2\left(x-1\right)}\)

\(=\dfrac{-8\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}\cdot\dfrac{x}{2\left(x-1\right)}\)

\(=\dfrac{-4x}{\left(x+4\right)\left(x-1\right)}\)

a: A=−4x/(x+4)(x−1)

1. ĐKXĐ: \(x\ne\pm1\)

2. \(A=\left(\dfrac{x+1}{x-1}-\dfrac{x+3}{x+1}\right)\cdot\dfrac{x+1}{2}\)

\(=\dfrac{\left(x+1\right)^2-\left(x-3\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{2}\)

\(=\dfrac{x^2+2x+1-x^2+4x-3}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{2}\)

\(=\dfrac{6x-2}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{2}\)

\(=\dfrac{2\left(x-3\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{x-3}{x-1}\)

3. Tại x = 5, A có giá trị là:

\(\dfrac{5-3}{5-1}=\dfrac{1}{2}\)

4. \(A=\dfrac{x-3}{x-1}\) \(=\dfrac{x-1-3}{x-1}=1-\dfrac{3}{x-1}\)

Để A nguyên => \(3⋮\left(x-1\right)\) hay \(\left(x-1\right)\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)

\(\Rightarrow\left\{{}\begin{matrix}x-1=1\\x-1=-1\\x-1=3\\x-1=-3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\left(tmđk\right)\\x=0\left(tmđk\right)\\x=4\left(tmđk\right)\\x=-2\left(tmđk\right)\end{matrix}\right.\)

Vậy: A nguyên khi \(x=\left\{2;0;4;-2\right\}\)

1: ĐKXĐ: x∉{1;-1}

2: \(A=\left(\frac{x+1}{x-1}-\frac{x+3}{x+1}\right)\cdot\frac{x+1}{2}\)

\(=\frac{\left(x+1\right)^2-\left(x+3\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\cdot\frac{x+1}{2}\)

\(=\frac{x^2+2x+1-\left(x^2+2x-3\right)}{x-1}\cdot\frac12=\frac{4}{2\left(x-1\right)}=\frac{2}{x-1}\)

3: THay x=5 vào A, ta được:

\(A=\frac{2}{5-1}=\frac24=\frac12\)

4: Để A là số nguyên thì 2⋮x-1

=>x-1∈{1;-1;2;-2}

=>x∈{2;0;3;-1}

Kết hợp ĐKXĐ, ta được: x∈{0;2;3}

\(x^2-4=0\)

=>\(x^2=4\)

=>x=2(loại) hoặc x=-2(nhận)

Khi x=-2 thì \(B=\frac{7}{x-2}=\frac{7}{-2-2}=\frac{7}{-4}=-\frac74\)