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

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

\(\Rightarrow\orbr{\begin{cases}2x^2\left(x-1\right)=2\left(x^2+1\right)\\2x^2\left(x-1\right)=-2\left(x^2+1\right)\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2\left(x-1\right)=x^2+1\\x^2\left(x-1\right)=-x^2-1\end{cases}}\)

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

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

Mk chỉ làm đc đến đây thôi còn lại mk chịu

đề bài ?

TÌM X,Y

  Vì {  a² + a ; a } và { b² + b ; b } bằng nhau nên ta có các trường hợp sau : 

 TH1 : a = b \( \implies\) a² +a = b² + b ( Luôn đúng )

 TH2 : a² + a = b và b² + b = a 

\( \implies\) a² + a + b² + b = a + b

\( \implies\) a² + b² = 0 ( 1 )

Ta có : a² \(\geq\) 0 ; b² \(\geq\) 0 \( \implies\) a² + b² \(\geq\) 0 ( 2 )

Từ ( 1 ) ; ( 2 ) Dấu " = " xảy ra \(\iff\) \(\hept{\begin{cases}a^2=0\\b^2=0\end{cases}}\) \(\iff\) \(\hept{\begin{cases}a=0\\b=0\end{cases}}\) \( \implies\) a = b = 0

KL : a = b

\(\frac{x}{z}=\frac{z}{y}\Leftrightarrow z^2=xy\)

Thay vào ta có: \(\frac{x^2+z^2}{z^2+y^2}=\frac{x^2+xy}{y^2+xy}=\frac{x\left(x+y\right)}{y\left(x+y\right)}=\frac{x}{y}\)

ta có x/z = z/ y 

=>  x^2/z^2 = z^2/ y^2 = xz/zy = x/y (1)

từ (1) adtcdtsbn 

 x^2/z^2 = z^2/ y^2 = \(\frac{x^2+z^2}{z^2+y^2}\)(2)

từ (1) (2) => đpcm

vậy................

x=4

y=1/2

Môn gì bạn

Môn j ???

lớp mấy

\(0,3x:\frac{10}{3}=\frac{6}{15}\)

\(\Leftrightarrow0,3x=\frac{3}{25}\)

\(\Leftrightarrow x=\frac{2}{5}\)

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