\(4x^2-4x+1+9y^2-6y+1+16z^2-8z+1=0\)

\(\Leftrightarrow\left(2x-1\right)^2+\left(3y-1\right)^2+\left(4z-1\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}2x-1=0\\3y-1=0\\4z-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{1}{3}\\x=\frac{1}{4}\end{cases}}\)

vay ................................................

Ta có : 

4x² + 9y² + 16z² - 4x - 6y - 8z + 3 = 0

( 2x ) ²  + ( 3y)² + ( 4z)² - 4x - 6y - 8z + 3 = 0

\([\left(2x\right)^2-2.2x+1]+[\left(3y\right)^2-2.3y+1]+[\left(4z\right)^2-2.4z+1]=0\)=0

( 2x-1)²  + ( 3y -1 )² + ( 4z - 1) ² = 0

Mà ( 2x-1)² \(\ge\)0 với mọi x

     ( 3y-1 )² \(\ge0\)với mọi y

      ( 4z - 1) ^{2 \(\ge0\)với mọi z} 

 nên \(\hept{\begin{cases}2x-1=0\\3y-1=0\\4z-1=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{1}{3}\\z=\frac{1}{4}\end{cases}}}\)

 Vậy x = 1/2 ; y = 1/3 ; z = 1/4 

\(x^2-2xy+y^2+x^2-10x+25=0\)

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

\(\Leftrightarrow\hept{\begin{cases}x-y=0\\x-5=0\end{cases}}\Leftrightarrow\hept{\begin{cases}y=5\\x=5\end{cases}}\)

vậy \(y=5\) va\(x=5\)

\(2x^2+y^2-2xy-10x+25=0\)

\(\Leftrightarrow(x^2-2xy+y^2)+\left(x^2-10x+25\right)=0\)

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

\(\Rightarrow\hept{\begin{cases}x-5=0\Rightarrow x=5\\x-y=5\Rightarrow y=5\end{cases}}\)

\(x^2+4x-9y^2+4\)

\(=\left(x^2+2.2x+2^2\right)-\left(3y\right)^2\)

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

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

\(x^2-9y^2-6y-1\)

\(=x^2-\left[\left(3y\right)^2+2.3y+1^2\right]\)

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

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

Tham khảo nhé~

a/Ta có:x²+4x-9y²+4

=x²+4x+4-(3y)²

=(x+2)²-(3y)²

=(x+2-3y)(x+2+3y)

b/Ta có:x²-9y²-6xy-1

=x²-6xy-(3y)²-1

=(x-3y-1)(x-3y+1)

\(2x^2=x\)

\(\Rightarrow2x^2-x=0\)

\(x\left(2x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\2x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}}\)

Vậy \(x=0\)hoặc \(x=\frac{1}{2}\)

\(x^3=x^5\)

\(\Rightarrow x^5-x^3=0\)

\(x^3.\left(x^2-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x^3=0\\x^2-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)

Vậy \(x=0\)hoặc \(x=1\)

\(x^2.\left(x+1\right)+2x\left(x+1\right)=0\)

\(\left(x+1\right)\left(x^2+2x\right)=0\)

\(x.\left(x+1\right)\left(x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\)hoặc \(x+2=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)hoặc \(x=-2\)

Vậy \(\orbr{\begin{cases}x=0\\x=-1\end{cases}}\) hoặc \(x=-2\)

\(x.\left(2x-3\right)-2\left(3-2x\right)=0\)

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

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

\(\Rightarrow\orbr{\begin{cases}2x-3=0\\x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=-2\end{cases}}}\)

Vậy \(x=\frac{3}{2}\)hoặc \(x=-2\)

\(2x^2-x=0\Leftrightarrow x\left(2x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{1}{2}\end{cases}}\)

\(S\left\{0;\frac{1}{2}\right\}\)

\(d)x^3-x^5=0\Leftrightarrow x^3\left(1-x^2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^3=0\\1-x^2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm\sqrt{1}\end{cases}}\)

\(S=\left\{0;\pm\sqrt{1}\right\}\)

các câu sau tương tự nha bn

Đặt \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)

   \(\Rightarrow2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+....+\frac{2}{n\left(n+1\right)\left(n+2\right)}\)

                \(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)

                \(=\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)

                \(=\frac{1}{2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)

                 \(=\frac{\left(n+1\right)\left(n+2\right)-2}{2\left(n+1\right)\left(n+2\right)}\)

\(\Rightarrow A=\frac{\left(n+1\right)\left(n+2\right)-2}{4\left(n+1\right)\left(n+2\right)}\)

TK nha!!

1)3x(x-2)=7(x-2)

<=>3x(x-2)-7(x-2)=0

<=>(x-2)(3x-7)=0

x-2=0=>x=2

3x-7=0=>x=7/3

cn lại lm tg tự

10)\(x^2-9x+20=0\)

\(\Leftrightarrow x^2-4x-5x+20=0\)

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

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

\(\Leftrightarrow\hept{\begin{cases}x=4\\x=5\end{cases}}\)