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Ta có: \(D=2x^2-8x-10\)

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

\(=2\left(x^2-4x+4-9\right)\)

\(=2\left(x-2\right)^2-18\ge-18\forall x\)

Dấu '=' xảy ra khi x=2

\(\Leftrightarrow\left(3x+7\right)\left(2x-5\right)=0\)

=>x=-7/3 hoặc x=5/2

8 tháng 1 2022

\(2x\left(3x+7\right)-15x-35=0\\ \Rightarrow2x\left(3x+7\right)-\left(15x+35\right)=0\\ \Rightarrow2x\left(3x+7\right)-5\left(3x+7\right)=0\\ \Rightarrow\left(2x-5\right)\left(3x+7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{3}\end{matrix}\right.\)

a: ta có: \(\left(8x+2\right)\left(1-3x\right)+\left(6x-1\right)\left(4x-10\right)=-50\)

\(\Leftrightarrow8x-24x^2+2-6x+24x^2-60x-4x+40=-50\)

\(\Leftrightarrow-62x=-92\)

hay \(x=\dfrac{46}{31}\)

b: ta có: \(\left(1-4x\right)\left(x-1\right)+4\left(3x+2\right)\left(x+3\right)=38\)

\(\Leftrightarrow x-1-4x^2+4x+4\left(3x^2+9x+2x+6\right)=38\)

\(\Leftrightarrow-4x^2+5x-1+12x^2+44x+24-38=0\)

\(\Leftrightarrow8x^2+49x-15=0\)

\(\text{Δ}=49^2-4\cdot8\cdot\left(-15\right)=2881\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là: 

\(\left\{{}\begin{matrix}x_1=\dfrac{-49-\sqrt{2881}}{16}\\x_2=\dfrac{-49+\sqrt{2881}}{16}\end{matrix}\right.\)

2 tháng 10 2021

bn ơi phần này làm áp dụng hằng đẳng thức đc k ạ

\(\Leftrightarrow4x^2-12x-4x^2+9=-3\)

=>-12x=-12

hay x=1

8 tháng 1 2022

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

\(4x^2-12x-4x^2+9+3=0\)

\(12-12x=0\\ \Rightarrow1-x=0\\ \Rightarrow x=1\)

8 tháng 1

\(\dfrac{1}{x-3}+\dfrac{3x^2-8x+10}{x^2-5x+6}-\dfrac{2x-4}{x-2}\left(ĐK:x\ne3;x\ne2\right)\)

\(=\dfrac{1}{x-3}+\dfrac{3x^2-8x+10}{x\left(x-2\right)-3\left(x-2\right)}-\dfrac{2x-4}{x-2}\)

\(=\dfrac{1}{x-3}+\dfrac{3x^2-8x+10}{\left(x-3\right)\left(x-2\right)}-\dfrac{2x-4}{x-2}\)

\(=\dfrac{x-2}{\left(x-2\right)\left(x-3\right)}+\dfrac{3x^2-8x+10}{\left(x-3\right)\left(x-2\right)}-\dfrac{\left(2x-4\right)\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{x-2+3x^2-8x+10-\left(2x^2-6x-4x+12\right)}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{3x^2-7x+8-2x^2+10x-12}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{x^2+3x-4}{\left(x-2\right)\left(x-3\right)}\)

\(=\dfrac{x^2+3x-4}{x^2-5x+6}\)

\(\left(a-\dfrac{a^2+b^2}{a+b}\right)\left(\dfrac{1}{b}+\dfrac{2}{a-b}\right)\)

\(=\dfrac{a^2+ab-a^2-b^2}{a+b}\cdot\dfrac{a-b+2b}{b\left(a-b\right)}\)

\(=\dfrac{b\left(a-b\right)}{a+b}\cdot\dfrac{a+b}{b\left(a-b\right)}\)

=1

\(\left(a-\dfrac{x^2+a^2}{x+a}\right)\left(\dfrac{2a}{x}-\dfrac{4a}{x-a}\right)\)

\(=\dfrac{ax+a^2-x^2-a^2}{x+a}\cdot\dfrac{2ax-2a^2-4ax}{x\left(x-a\right)}\)

\(=\dfrac{a\left(x-a\right)}{x+a}\cdot\dfrac{-2a^2-2ax}{x\left(x-a\right)}\)

\(=\dfrac{a}{x+a}\cdot\dfrac{-2a\left(a+x\right)}{x}\)

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

7 tháng 9 2021

\(\left(a-\dfrac{x^2+a^2}{x+a}\right)\left(\dfrac{2a}{x}-\dfrac{4a}{x-a}\right)\)

\(=\dfrac{ax+a^2-x^2-a^2}{x+a}.\dfrac{2a\left(x-a\right)-4ax}{x\left(x-a\right)}\)

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

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

\(=2a\)