câu 1 B 

câu 2 D

câu 3 ko bt 

câu 4 x=-1/2; x = -(căn bậc hai(3)*i-1)/4;x = (căn bậc hai(3)*i+1)/4;

câu 5 x=-5/3, x=0, x=1

Câu 1:  x² + 2 xy + y²   bằng:

A. x² + y²                   B.(x + y)²                  C. y² – x²                  D. x² – y²

Câu 2:  (4x + 2)(4x – 2)  bằng:

A. 4x² + 4                  B. 4x² – 4                 C. 16x² + 4                D. 16x² – 4

Câu 3: 25a²  + 9b²  - 30ab  bằng:

A.(5a-9b)²                  B.(5a – 3b)²              C.(5a+3b)²                D.(5a)² – (3b)²

Câu 4: 8x³ +1 bằng

A.(2x+1).(4x²-2x+1)      B. (2x-1).(4x²+2x+1)       C.(2x+1)³            D.(2x)³-1³

Câu 5:Thực hiện phép nhân  x(3x² + 2x - 5) ta được:

A.3x³ - 2x² – 5x          B. 3x³ + 2x² – 5x      C. 3x³ - 2x² +5x         D. 3x³ + 2x² + 5x

a: \(\dfrac{x+10}{4x-8}\cdot\dfrac{4-2x}{x+2}\)

\(=\dfrac{x+10}{4\left(x-2\right)}\cdot\dfrac{-2\left(x-2\right)}{x+2}=\dfrac{-\left(x+10\right)}{2\left(x+2\right)}\)

b: \(\dfrac{1-4x^2}{x^2+4x}:\dfrac{2-4x}{3x}\)

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

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

c: \(=\dfrac{4y^2}{7x^4}\cdot\dfrac{35x^2}{-8y}=\dfrac{5}{x^2}\cdot\dfrac{-1}{2}\cdot y=\dfrac{-5y}{2x^2}\)

d: \(=\dfrac{\left(x-2\right)\left(x+2\right)}{3\left(x+4\right)}\cdot\dfrac{x+4}{2\left(x-2\right)}=\dfrac{x+2}{6}\)

thứ nhất bn đăng sai môn 

thứ hai bn giải r đăng lmj :???

Thứ nhất đang sai môn 

Thứ hai không biết giải fndf]-0jhdfuhiofghjfgoihjfgopihjfgihjohjgo;hjghghgdjhldhjdfighjs;dligjlkdfgjdhfghfgh41fg6j541fg3j5h4gf6j54dgh65gf4654j

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a) \(\sqrt[]{x^2-2x+4}=2x-2\)

\(\Leftrightarrow\sqrt[]{x^2-2x+4}=2\left(x-1\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}2\left(x-1\right)\ge0\\x^2-2x+4=4\left(x-1\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1\ge0\\x^2-2x+4=4x^2-8x+4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\3x^2-6x=0\end{matrix}\right.\) \(\left(1\right)\)

Giải pt \(3x^2-6x=0\)

\(\Leftrightarrow3x\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=2\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow x=2\)

c) \(\sqrt{x^2-3x+2}=\sqrt[]{x-1}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-1\ge0\\x^2-3x+2=x-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x^2-4x+3=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\x=1\cup x=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)

a: \(=\dfrac{\left|x+2\right|}{x-1}\)

b: \(=x-2y-\left|x-2y\right|\)\(=\left[{}\begin{matrix}x-2y-x+2y=0\\x-2y+x-2y=2x-4y\end{matrix}\right.\)

c: \(=\dfrac{\left|x+2\right|}{\left(x+2\right)\left(x-2\right)}=\pm\dfrac{1}{x-2}\)

\(a/\)

\(4x-4y+x^2-2xy+y^2\)

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

\(=4\left(x-y\right)+\left(x-y\right)^2\)

\(=\left(x-y\right)\left(4+x-y\right)\)

\(b/\)

\(x^4-4x^3-8x^2+8x\)

\(=\left(x^4+8x\right)-\left(4x^3+8x^2\right)\)

\(=x\left(x^3+8\right)-4x^2\left(x+2\right)\)

\(=x\left(x+2\right)\left(x^2-2x+4\right)-4x^2\left(x+2\right)\)

\(=x\left(x+2\right)\left(x^2-2x+4-4x\right)\)

\(=x\left(x+2\right)\left(x^2-6x-4\right)\)

\(d/\)

\(x^4-x^2+2x-1\)

\(=x^4-\left(x-1\right)^2\)

\(=\left(x^2+x-1\right)\left(x^2-x+1\right)\)

\(e/\)(Xem lại đề)

\(x^4+x^3+x^2+2x+1\)

\(=\left(x^4+x^3\right)+\left(x^2+2x+1\right)\)

\(=x^3\left(x+1\right)+\left(x+1\right)^2\)

\(=\left(x+1\right)\left(x^3+x+1\right)\)

\(f/\)

\(x^3-4x^2+4x-1\)

\(=x\left(x^2-4x+4\right)-1^2\)

\(=x\left(x-2\right)^2-1\)

\(=[\sqrt{x}\left(x-2\right)]^2-1\)

\(=[\sqrt{x}\left(x-2\right)-1][\sqrt{x}\left(x-2\right)+1]\)

\(c/\)

\(x^3+x^2-4x-4\)

\(=\left(x^3-2x^2\right)+\left(3x^2-6x\right)+\left(2x-4\right)\)

\(=x^2\left(x-2\right)+3x\left(x-2\right)+2\left(x-2\right)\)

\(=\left(x-2\right)\left(x^2+3x+2\right)\)

\(=\left(x-2\right)[\left(x^2+x\right)+\left(2x+2\right)]\)

\(=\left(x-2\right)\left(x+1\right)\left(x+2\right)\)

d) \(\sqrt{x^2-6x+9}=2\Leftrightarrow\sqrt{\left(x-3\right)^2}=2\Leftrightarrow x-3=2\Leftrightarrow x=5\)

e) đk: \(x\ge2\)\(\sqrt{x^2-3x+2}=\sqrt{x-1}\Leftrightarrow\sqrt{\left(x-2\right)\left(x-1\right)}=\sqrt{x-1}\Leftrightarrow\sqrt{x-2}=1\Leftrightarrow x-2=1\Leftrightarrow x=3\)f) \(\sqrt{4x^2-4x+1}=\sqrt{x^2-6x+9}\Leftrightarrow\sqrt{\left(2x-1\right)^2}=\sqrt{\left(x-3\right)^2}\Leftrightarrow2x-1=x-3\Leftrightarrow x=-2\)

c: Ta có: \(\sqrt{x+4\sqrt{x-4}}=2\)

\(\Leftrightarrow\left|\sqrt{x-4}+2\right|=2\)

\(\Leftrightarrow x-4=0\)

hay x=4

a) \(\sqrt{x-1+2\sqrt{x-1}.1+1^2}=2;đk:x\)≥1
⇔\(\sqrt{\left(\sqrt{x-1}\right)^2+2\sqrt{x-1}.1+1^2}=2\left(hđt-1\right)\)
⇔\(\sqrt{\left(\sqrt{x-1}+1\right)^2=2}\)
⇔|\(\sqrt{x-1}+1\)|=2
⇔\(\left[{}\begin{matrix}\sqrt{x+1}-1=2\\\sqrt{x+1-1}=-2\end{matrix}\right.\)⇔\(\left[{}\begin{matrix}\sqrt{x+1}=3\\\sqrt{x+1}=-1\left(L\right)\end{matrix}\right.\)⇔x+1=9⇔x=10(TM)
→S={10}

Câu d đề có đúng ko bn

mk thấy hơi sai Nguyen Thi tuong Vi