98

98

98

\(a,\left|2x+\dfrac{1}{2}\right|=0\\ \Leftrightarrow2x+\dfrac{1}{2}=0\\ \Leftrightarrow2x=-\dfrac{1}{2}\\ \Leftrightarrow x=-\dfrac{1}{4}\\ b,\left|3x+\dfrac{3}{4}\right|=3\\ \Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{3}{4}=3\\3x+\dfrac{3}{4}=-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=\dfrac{9}{4}\\3x=-\dfrac{15}{4}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{5}{4}\end{matrix}\right.\)

-2

-19/3

 k mình nha

\(\dfrac{-11}{4}-\dfrac{6}{8}=\dfrac{-11}{4}-\dfrac{3}{4}=\dfrac{-14}{4}=\dfrac{-7}{2}\)

\(\dfrac{-7}{1}-\dfrac{2}{3}=\dfrac{-21}{3}-\dfrac{2}{3}=\dfrac{-19}{3}\)

\(3x^2-3+2x^4+6x^2-4-5x-8x^3=2x^4-8x^3-3x^2-5x-7\)

\(x^2+4x+5=2\sqrt{2x+3}\)

\(ĐK:x\ge-\dfrac{3}{2}\)

\(pt\Leftrightarrow(2x+3-2\sqrt{2x+3}+1)+x^2+2x+1=0\)

\(\Leftrightarrow\left(\sqrt{2x+3}-1\right)^2=-\left(x+1\right)^2\)

Vì \(\left(\sqrt{2x+3}-1\right)^2\ge0;-\left(x+1\right)^2\le0\forall x\)

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

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

\(\Leftrightarrow x=-1\left(tm\right)\)

Vậy, pt có nghiệm duy nhất là x=-1

Giải pt à bạn

Bài 1:

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

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

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

\(=0\)

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

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

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

\(=2x^2+4\)

c) \(\left(3-x\right)^2+2\left(x-3\right)\left(x+7\right)+\left(x+7\right)^2\)

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

\(=\left[\left(x-3\right)+\left(x+7\right)\right]^2\)

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

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

Bài 2:

a) \(x^2-4-4\left(2-x\right)\)

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

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

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

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

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

b) \(x^3+8\left(1-2x\right)^3\)

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

\(=\left[x+2\left(1-2x\right)\right]\left[x^2-x\cdot2\left(1-2x\right)+2^2\left(1-2x\right)^2\right]\)

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

\(=\left(2-3x\right)\left(x^2-2x+4x^2+4-16x+16x^2\right)\)

\(=\left(2-3x\right)\left(21x^2-18x+4\right)\)

Theo mình thì câu b có lẽ là tổng của 3 bình phương.

\(\left(x-5\right)^2=\left(18\dfrac{1}{3}:5\right).\dfrac{11}{3}\)

\(\Leftrightarrow\left(x-5\right)^2=\dfrac{55}{3}.\dfrac{1}{5}.\dfrac{11}{3}\)

\(\Leftrightarrow\left(x-5\right)^2=\dfrac{121}{9}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=\dfrac{11}{3}\\x-5=-\dfrac{11}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{26}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\)

\(=-7\cdot\left[16+\left(-36\right):\left(-9\right)\right]+125=-77\cdot20+125=-140+125=-15\)

\(-7.\left[\left(-2\right)^4+\left(-36\right):\left(-3\right)^2\right]-\left(-5\right)^3\\ =-7.\left[16+\left(-36\right):9\right]+125\\ =-7.\left[16+\left(-4\right)\right]+125\\ =-7.12+125\\ =-84+125\\ =41\)