là đấy

kết quả là2

|x²+|3x-1||=x²+7

Vì |x²+|3x-1|| \(\ge\) 0;

    x²+7 \(\ge\) 0

nên VT=VP

<=>x²+|3x-1|=x²+7

<=>|3x-1|=x²+7-x²=7

<=>\(\int^{3x-1=7\Rightarrow3x=8\Rightarrow x=\frac{8}{3}}_{3x-1=-7\Rightarrow3x=-6\Rightarrow x=-2}\)

Vậy x \(\in\) (-2;8/3}

\(\Rightarrow x=4,5\)

\(6x^2-\left(2x+5\right)\left(3x-2\right)=7.\)

\(6x^2-\left(6x^2-4x+15x-10\right)=7\)

\(6x^2-6x^2+4x-15x+10=7\)

\(-11x+10=7\)

\(-11x=-3\)

\(x=\frac{-3}{-11}=\frac{3}{11}\)

\(6x^2-\left(2x+5\right)\left(3x-2\right)=7\)

\(6x^2-\left(6x^2-4x+15x-10\right)=7\)

+ Áp dụng quy tắc bở ngoặc ta có :

\(6x^2-6x^2+4x-15x+10=7\)

\(-11x+10=7\)

        \(-11x=7-10\)

            \(-11x=-3\)

                        \(x=-3:-11\)

                         \(x=\frac{-3}{-11}\)

                       \(\Rightarrow x=\frac{3}{11}\)

     Vậy \(x=\frac{3}{11}\)

Chúc bạn học tốt !!!

(5x-2)(3x+1)+(7-15x)(x+3)=-20

<=> 15x²+5x-6x-2+7x+21-15x²-45x+20=0

<=>39-39x=0

<=>39(1-x)=0

<=>1-x=0

=>x=1

(5x-2)(3x+1)+(7-15x)(x+3)=-20

=>\(15x^2-6x+5x-2+7x-15^2+21-45x=-20\)

=>\(-39x+19=-20\)

=>\(-39x=-39\)

=>\(x=1\)

vậy x=1

(5x-2)(3x+1)+(7-15x)(x+3)=-20

<=>\(15x^2+5x-6x-2+7x-15x^2+21-45x=-20\)

<=>\(-39x+19=0\)

<=>\(-39x=-19\)

<=>\(x=\dfrac{19}{39}\)

Vậy \(x=\dfrac{19}{39}\)

\(25\%x-\frac{7}{8}+x=\frac{2}{3}x\)

<=> \(\frac{1}{4}x-\frac{7}{8}+1x=\frac{2}{3}x\)

<=> \(\frac{1}{4}x+1x-\frac{2}{3}x=\frac{7}{8}\)

<=> \(x\left(\frac{1}{4}+1-\frac{2}{3}\right)=\frac{7}{8}\)

<=> \(x\cdot\frac{7}{12}=\frac{7}{8}\)

<=> \(x=\frac{12}{8}=\frac{3}{2}\)

Bài làm:

Ta có: \(25\%x-\frac{7}{8}+x=\frac{2}{3}x\)

\(\Leftrightarrow\frac{x}{4}+x-\frac{2}{3}x=\frac{7}{8}\)

\(\Leftrightarrow\frac{7}{12}x=\frac{7}{8}\)

\(\Leftrightarrow x=\frac{3}{2}\)