Ko ghi lại đề nha em gái!

\(\Leftrightarrow\orbr{\begin{cases}x:\left(-\frac{5}{6}\right)-\frac{3}{4}=2\\x:\left(-\frac{5}{6}\right)-\frac{3}{4}=-2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x.\left(-\frac{6}{5}\right)=2+\frac{3}{4}\\x.\left(-\frac{6}{5}\right)=-2+\frac{3}{4}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}-\frac{6}{5}x=\frac{11}{4}\\-\frac{6}{5}x=-\frac{5}{4}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{55}{24}\\x=\frac{25}{24}\end{cases}}\)

Vậy x1 = -55/24 ; x2 = 25/24 

\(\left|x:\left(-\frac{5}{6}\right)-\frac{3}{4}\right|=2\)

\(\Leftrightarrow\orbr{\begin{cases}x:\left(-\frac{5}{6}\right)-\frac{3}{4}=2\\x:\left(-\frac{5}{6}\right)-\frac{3}{4}=-2\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x.\frac{-6}{5}=2+\frac{3}{4}\\x.\frac{-6}{5}=-2+\frac{3}{4}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x.\frac{-6}{5}=\frac{11}{4}\\x.\frac{-6}{5}=-\frac{5}{4}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{55}{24}\\x=\frac{25}{24}\end{cases}}\)

ta có : a(b+1)(a+1)=(a+1)(b+1)     (1)  

chia cả 2 vế cho (a+1)(b+1) => (1) <=> a=1

thay a =1 vào (1) => b=1

=> a.b=1

\(A=m\left(3m-2m^2-7\right)-\left(5-m\right)\left(m^2+3\right)+\left(m^2-m-14\right)\)

\(=3m^2-2m^3-7m-\left(5m^2-m^3+15-3m\right)+m^2-m-14\)

\(=4m^2-2m^3-8m-5m^2+m^3-15+3m-14\)

\(=-m^2-m^3-5m-29\)

\(\sqrt{2}A=\sqrt{2}\sqrt{13+\sqrt{2}+5\sqrt{1+2\sqrt{2}}}+\sqrt{2}\sqrt{13+\sqrt{2}-5\sqrt{1+2\sqrt{2}}}\)

\(=\sqrt{26+2\sqrt{2}+5.2\sqrt{1+2\sqrt{2}}}+\sqrt{26+2\sqrt{2}-5.2\sqrt{1+2\sqrt{2}}}\)

\(=\sqrt{5^2+2.5.\sqrt{1+2\sqrt{2}}+\left(1+2\sqrt{2}\right)}+\sqrt{5^2-2.5.\sqrt{1+2\sqrt{2}}+\left(1+2\sqrt{2}\right)}\)

\(=\sqrt{\left(\sqrt{1+2\sqrt{2}}+5\right)^2}+\sqrt{\left(\sqrt{1+2\sqrt{2}}-5\right)^2}\)

\(=\left|\sqrt{1+2\sqrt{2}}+5\right|+\left|\sqrt{1+2\sqrt{2}}-5\right|\)

\(=\sqrt{1+2\sqrt{2}}+5+5-\sqrt{1+2\sqrt{2}}=10\)

=> \(A=\frac{10}{\sqrt{2}}=5\sqrt{2}\)

Ta có:x:y:z=3:4:5

=>\(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\)

Mà 2x²+2y²-3z²=-100

=>\(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}=\frac{2x^2+2y^2-3z^2}{18+32-75}=\frac{-100}{-25}=4\)

=>x²=4x3=12=>x=\(\sqrt{12}\)

y²=4x4=16=>x=4

z²=4x5=20=>x=\(\sqrt{20}\)

Vậy,ta có x=\(\sqrt{12}\) y=4                 z=\(\sqrt{20}\)

Xin lỗi bạn mình làm sai mình sẽ làm lại

a, Ư(12)=\(\left\{\pm1;\pm2;\pm3;\pm4\pm6\pm12\right\}\)

Ư(18)=\(\left\{\pm1;\pm2;\pm3;\pm6;\pm9;\pm18\right\}\)

ƯC(12;18)=\(\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

c,ƯC(45;20;72)=\(\left\{\pm1\right\}\)

Câu hỏi của Công Chúa Họ Kim - Toán lớp 6 - Học toán với OnlineMath

<=> 7x - 42 = 4x + 9 

<=> (7 - 4 )x = 9 +42

<=>    3x       =   51 

<=>      x       = 17 

\(7\left(x-6\right)=4x+9\)\(\)

\(\Rightarrow7x-42-4x-9=0\)

\(\Rightarrow3x=51\)

\(\Rightarrow x=17\)

Vậy x =17

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