Ta có : \(\left|x+\frac{13}{14}\right|=-\left|x-\frac{3}{7}\right|\)

\(\Rightarrow\left|x+\frac{13}{14}\right|+\left|x-\frac{3}{7}\right|=0\)

Mà : \(\left|x+\frac{13}{14}\right|\ge0\forall x\)

      \(\left|x-\frac{3}{7}\right|\ge0\forall x\)

Nên : \(\orbr{\begin{cases}\left|x+\frac{13}{14}\right|=0\\\left|x-\frac{3}{7}\right|=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x+\frac{13}{14}=0\\x-\frac{3}{7}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{13}{14}\\x=\frac{3}{7}\end{cases}}\)

\(\left(2x-1\right)^2-3.\left(x+2\right)^2=4.\left(x-2\right)-5.\left(x-1\right)^2\)

\(\Leftrightarrow4x^2-4x+1-3\left(x^2+4x+4\right)=4x-8-5.\left(x^2-2x+1\right)\)

\(\Leftrightarrow4x^2-4x+1-3x^2-7x-12=4x-8-5x^2+10x-5\)

\(\Leftrightarrow x^2-11x-11=14x-13-5x^2\)

\(\Leftrightarrow6x^2-25x+2=0\)

Tự làm tiếp nha

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bạn giải tiếp giúp mk với được ko

Giải tới đây pt có 2 ngiệm\(\hept{\begin{cases}x_1=\frac{25+\sqrt{577}}{12}\\x_2=\frac{25-\sqrt{577}}{12}\end{cases}}\)

Ta có: \(D=\left(x-y\right)^2+2\left(x^2-y^2\right)+\left(x+y\right)^2\)

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

\(=4x^2\)

b)(2x - 1)^2 - (2x + 5) (2x - 5 ) = 18

4x 2 -4x+1-4x 2+25=18

26-4x=18

4x=8

x=2

a,27x-18=2x-3x^2

<=> 3x^2-2x+27-18x=0

<=> 3x^2-20x+27=0

\(\Delta\)= 20^2-4-12.27

tính \(\Delta\)rồi tìm x1 ,x2

â)\(9\left(3x-2\right)=x\left(2-3x\right)\)

\(\Leftrightarrow27x-18=2x-3x^2\)

\(\Leftrightarrow27x-18-2x+3x^2=0\)

\(\Leftrightarrow3x^2+25x-18=0\)

\(\Leftrightarrow3x^2+27x-2x-18=0\)

\(\Leftrightarrow\left(3x-2\right)\left(x+9\right)=0\)

       \(\Rightarrow\orbr{\begin{cases}3x-2=0\\x+9=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{2}{3}\\x=-9\end{cases}}\)

b)\(\left(2x-1\right)^2-\left(2x+5\right)\left(2x-5\right)=18\)

\(\Leftrightarrow4x^2-4x+1-4x^2+25=18\)

\(\Leftrightarrow26-4x=18\)

 \(\Leftrightarrow4x=8\)

      \(\Rightarrow x=2\)

c)\(5x\left(x-5\right)-2x+10=0\)

\(\Leftrightarrow5x^2-10x-2x+10=0\)

\(\Leftrightarrow5x^2-12x+10=0\)

\(\Leftrightarrow x^2-6x+2=0\)

\(\Leftrightarrow x^2-6x+9-7=0\)

\(\Leftrightarrow\left(x-3\right)^2=7\)

       \(\Rightarrow\orbr{\begin{cases}x-3=\sqrt{7}\\x-3=-\sqrt{7}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\sqrt{7}+3\\x=-\sqrt{7}+3\end{cases}}\)

d)\(x^2-5=0\)

\(\Leftrightarrow x^2=5\)

    \(\Rightarrow x=\sqrt{5};-\sqrt{5}\)

e)\(x^3+5x^2-4x-20=0\)

\(\Leftrightarrow x^3-2x^2+7x^2-14x+10x-20=0\)

\(\Leftrightarrow x^2\left(x-2\right)+7x\left(x-2\right)+10\left(x-2\right)=0\)

\(\Leftrightarrow\left(x^2+7x+10\right)\left(x-2\right)=0\)

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

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

      \(\Rightarrow\orbr{\begin{cases}x+5=0\\x^2-4=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-5\\x^2=4\end{cases}\Rightarrow}\orbr{\begin{cases}x=-5\\x=-2;2\end{cases}}\)

a) \(x^2+\left(y-\dfrac{1}{10}\right)^4=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y-\dfrac{1}{10}=0\end{matrix}\right.\)( do \(x^2\ge0,\left(y-\dfrac{1}{10}\right)^4\ge0\))

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\dfrac{1}{10}\end{matrix}\right.\)

b) \(\left(\dfrac{1}{2}.x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\le0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x-5=0\\y^2-\dfrac{1}{4}=0\end{matrix}\right.\)( do \(\left(\dfrac{1}{2}x-5\right)^{20}\ge0,\left(y^2-\dfrac{1}{4}\right)^{10}\ge0\))

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x=5\\y^2=\dfrac{1}{4}\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=\pm\dfrac{1}{2}\end{matrix}\right.\)

\(a,\Leftrightarrow\left\{{}\begin{matrix}x=0\\y-\dfrac{1}{10}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\dfrac{1}{10}\end{matrix}\right.\\ b,\left\{{}\begin{matrix}\left(\dfrac{1}{2}x-5\right)^{20}\ge0\\\left(y^2-\dfrac{1}{4}\right)^{10}\ge0\end{matrix}\right.\Leftrightarrow\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\ge0\)

Mà \(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\le0\)

\(\Leftrightarrow\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}=0\\ \Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x=5\\y^2=\dfrac{1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=10\\y=\pm\dfrac{1}{2}\end{matrix}\right.\)

a) \(x^2+\left(y-\dfrac{1}{10}\right)^4=0\)

Mà \(x^2+\left(y-\dfrac{1}{10}\right)^4\ge0\forall x;y\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=0\\\left(y-\dfrac{1}{10}\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\dfrac{1}{10}\end{matrix}\right.\)

Vậy \(\left(x;y\right)=\left(0;\dfrac{1}{10}\right)\)

b) \(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\le0\)

Mà \(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{10}\ge0\forall x;y\)

\(\Rightarrow\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}\left(\dfrac{1}{2}x-5\right)^{20}=0\\\left(y^2-\dfrac{1}{4}\right)^{10}=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=10\\\left[{}\begin{matrix}y=\dfrac{1}{2}\\y=-\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\)

Vậy \(\left(x;y\right)\in\left\{\left(10;\dfrac{1}{2}\right);\left(10;-\dfrac{1}{2}\right)\right\}\)