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) \(\sqrt{x^2-3x+2}=\sqrt[]{x-1}\)

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

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

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

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

ối giồi giải mấy bài trẻ con này thì chắc ko bao giờ vươn ra thế giới, há há

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

não lớp 5 (me) khi nhìn bài này kiểu : (banh não)

dạng toán của lớp 1 à 

không biết ??..!

a,sửa đề :  \(\left(\frac{1}{x^2+4x+4}-\frac{1}{x^2-4x+4}\right):\left(\frac{1}{x+2}+\frac{1}{x^2-4}\right)\)

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

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

\(=\frac{-8x\left(x+2\right)\left(x-2\right)}{\left(x+2\right)^2\left(x-2\right)^2\left(x-1\right)}=\frac{-8x}{\left(x-1\right)\left(x^2-4\right)}\)

b, \(\left(\frac{2x}{2x-y}-\frac{4x^2}{4x^2+4xy+y^2}\right):\left(\frac{2x}{4x^2-y^2}+\frac{1}{y-2x}\right)\)

\(=\left(\frac{2x}{2x-y}-\frac{4x^2}{\left(2x+y\right)^2}\right):\left(\frac{2x}{\left(2x-y\right)\left(2x+y\right)}-\frac{1}{2x-y}\right)\)

\(=\left(\frac{2x\left(2x+y\right)^2-4x^2\left(2x-y\right)}{\left(2x-y\right)\left(2x+y\right)^2}\right):\left(\frac{2x-\left(2x+y\right)}{\left(2x-y\right)\left(2x+y\right)}\right)\)

\(=\left(\frac{8x^3+8x^2y+2xy^2-8x^3+4x^2y}{\left(2x-y\right)\left(2x+y\right)^2}\right):\left(\frac{-y}{\left(2x-y\right)\left(2x+y\right)}\right)\)

\(=-\left(\frac{12x^2y+xy^2}{2x+y}\right)=\frac{-12x^2y-xy^2}{2x+y}\)

a) x^2+4x+3=x^2+x+3x+3=x(x+1)+3(x+1)=(x+1)(x+3)

b) 4x^2+4x-3=4x^2+4x+1-4=(2x+1)^2-4=(2x+1-2)(2x+1+2)=(2x-1)(2x+3)

c) x^2-x-12=x^2-4x+3x-12=x(x-4)+3(x-4)=(x-4)(x+3)

d) 4x^4+4x^2y^2-8y^4=4(x^4+x^2y^2-2y^4)=4(x^4-x^2y^2+2x^2y^2-2y^4)=4(x^2-y^2)(x^2+2y^2)=4(x-y)(x+y)(x^2+2y^2)

a) \(x^2+4x+3\)

\(=x^2+x+3x+3\)

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

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

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

c) \(x^2-x-12\)

\(=x^2-4x+3x-12\)

\(=\left(x^2-4x\right)+\left(3x-12\right)\)

\(=x\left(x-4\right)+3\left(x-4\right)\)

\(=\left(x-4\right)\left(x+3\right)\)

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

\(=x^2+x+3x+3\)

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

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

a. \(\dfrac{\left(x-2\right)^2}{2x\left(x-2\right)}=\dfrac{x-2}{2x}\)

b. \(\dfrac{\dfrac{1}{2}-2}{2.\dfrac{1}{2}}=-1,5\)

Bạn ghi lại đề đi, khó nhìn quá

a) Ta có: \(\left(x-2\right)\cdot x=2x\cdot\left(x+5\right)\)

\(\Leftrightarrow x\cdot\left(x-2\right)-2x\left(x+5\right)=0\)

\(\Leftrightarrow x\cdot\left[x-2-2\left(x+5\right)\right]=0\)

\(\Leftrightarrow x\left(x-2-2x-10\right)=0\)

\(\Leftrightarrow x\left(-x-8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\-x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\-x=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\)

Vậy: S={0;-8}

b) Ta có: \(\left(2x-5\right)\left(x+11\right)=\left(5-2x\right)\left(2x+1\right)\)

\(\Leftrightarrow\left(2x-5\right)\left(x+11\right)-\left(5-2x\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(x+11\right)+\left(2x-5\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(x+11+2x+1\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(3x+12\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\3x+12=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\3x=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-4\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{5}{2};-4\right\}\)

c) Ta có: \(x^2+6x+9=4x^2\)

\(\Leftrightarrow\left(x+3\right)^2-\left(2x\right)^2=0\)

\(\Leftrightarrow\left(x+3-2x\right)\left(x+3+2x\right)=0\)

\(\Leftrightarrow\left(-x+3\right)\left(3x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-x+3=0\\3x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=-3\\3x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

Vậy: S={3;-1}

d) Ta có: \(\left(x+2\right)\left(5-4x\right)=x^2+4x+4\)

\(\Leftrightarrow\left(x+2\right)\left(5-4x\right)-\left(x^2+4x+4\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(5-4x\right)-\left(x+2\right)^2=0\)

\(\Leftrightarrow\left(x+2\right)\left(5-4x-x-2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(-5x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\-5x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\-5x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{5}\end{matrix}\right.\)

Vậy: \(S=\left\{-2;\dfrac{3}{5}\right\}\)