a.
\(\left|x+10\right|=15\Rightarrow\orbr{\begin{cases}x+10=15\\x+10=-15\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-25\end{cases}}}\)
b.
\(\left|x-3\right|+5=7\Rightarrow\left|x-3\right|=2\Rightarrow\orbr{\begin{cases}x-3=2\\x-3=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=1\end{cases}}\)
c.
\(\left|x-3\right|+12=6\Rightarrow\left|x-3\right|=-6\Rightarrow x=\Phi\)
Phương trình vô nghiệm
d.
\(\left(2x+4\right)\left(3x-9\right)=0\Rightarrow\orbr{\begin{cases}2x+4=0\\3x-9=0\end{cases}\Rightarrow}\orbr{\begin{cases}2x=-4\\3x=9\end{cases}}\Rightarrow\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)
e.
\(x^2-5x=0\Rightarrow x\left(x-5\right)=0\Rightarrow\orbr{\begin{cases}x=0\\x-5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}\)
f.
\(\left(x+3\right)\left(4-2x\right)=70\Rightarrow4x-2x^2+7-6x=70\Rightarrow2x^2+2x+63=0\Rightarrow2\left(x+\frac{1}{2}\right)^2+\frac{123}{2}=0\)(vô lí)
Vậy phương trình vô nghiệm 

 

1,22

\(\dfrac{6}{7}\cdot\dfrac{8}{9}+\dfrac{7}{9}\cdot\dfrac{2}{7}-\dfrac{18}{21}\cdot\dfrac{1}{9}+\dfrac{7}{9}\cdot\dfrac{3}{7}\)

\(=\dfrac{6}{7}\cdot\dfrac{8}{9}-\dfrac{6}{7}\cdot\dfrac{1}{9}+\dfrac{7}{9}\left(\dfrac{2}{7}+\dfrac{3}{7}\right)\)

\(=\dfrac{6}{7}\left(\dfrac{8}{9}-\dfrac{1}{9}\right)+\dfrac{7}{9}\cdot\dfrac{5}{7}\)

\(=\dfrac{7}{9}\left(\dfrac{6}{7}+\dfrac{5}{7}\right)=\dfrac{7}{9}\cdot\dfrac{11}{7}=\dfrac{11}{9}\)

c: ĐKXĐ: x<>-1

Để C là số nguyên thì \(2x-7⋮x+1\)

=>\(2x+2-9⋮x+1\)

=>\(x+1\in\left\{1;-1;3;-3;9;-9\right\}\)

=>\(x\in\left\{0;-2;2;-4;8;-10\right\}\)

d: ĐKXĐ: x<>-3

Để D là số nguyên thì \(5x+9⋮x+3\)

=>\(5x+15-6⋮x+3\)

=>\(x+3\inƯ\left(-6\right)\)

=>\(x+3\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)

=>\(x\in\left\{-2;-4;-1;-5;0;-6;3;-9\right\}\)

Đề bài là gì thế bạn?

\(\frac{1}{2}\left(\frac{4}{9}-x\right)-\frac{3}{2}\left(16-x\right)+\frac{1}{2}\left(5x+10\right)=0\)

\(\Leftrightarrow\frac{2}{9}-\frac{1}{2}x-24+\frac{3}{2}x+\frac{5}{2}x+5=0\)

\(\Leftrightarrow-\frac{169}{9}=\frac{7}{2}x\Leftrightarrow x=-\frac{338}{63}\)

Sai thì thông cảm cho mk nha

\(5x+7⋮x^2\)

\(\Rightarrow\left(5x+7\right)\left(5x-7\right)⋮x^2\)

\(25x^2-49⋮x^2\)

\(49⋮x^2\)

\(x^2\inƯ\left(49\right)\)

\(x^2\in\left\{1;49\right\}\) vì x² là số chính phương và x² \(\ge\)0

\(x\in\left\{1;-1;7;-7\right\}\)

Thay vào ta được các giá trị x thỏa mãn là : .....( bạn tự liệt kê ra nhé )

\(6x+4⋮2x-1\)

\(3\left(2x-1\right)+7⋮2x-1\)

\(7⋮2x-1\)

\(2x-1\inƯ\left(7\right)\)

\(2x-1\in\left\{1;7;-1;-7\right\}\)

\(2x\in\left\{2;8;0;-6\right\}\)

\(x\in\left\{1;4;0;-3\right\}\)

Vậy .........................................................................................................................

\(a,-\frac{9}{12}=-\frac{9:3}{12:3}=-\frac{3}{4}\)

\(\frac{-18}{-24}=\frac{\left(-18\right):\left(-6\right)}{\left(-24\right):\left(-6\right)}=\frac{3}{4}\)

\(-\frac{35}{70}=-\frac{35:35}{70:35}=\frac{1}{2}\)

\(-\frac{9}{27}=-\frac{9:9}{27:9}=-\frac{1}{3}\)

\(b,\frac{1313}{4242}=\frac{1313:101}{4242:101}=\frac{13}{42}\)

\(\frac{-353535}{-424242}=\frac{\left(-353535\right):\left(-70707\right)}{\left(-424242\right):\left(-70707\right)}=\frac{5}{6}\)

\(c,\frac{2^3\times4^3\times5^4}{8^2\times25^3\times7}=\frac{2^3\times4^3\times25^2}{8\times8^2\times25^2\times25\times7}\)         ( 4^3 = 8^2 ; 5^4 = 25^2 )

\(=\frac{1}{25\times7}=\frac{1}{175}\)

\(a.\frac{-9}{-12}=\frac{-9:3}{-12:3}=\frac{-3}{-4}.\)

\(\frac{-18}{-24}=\frac{-18:6}{-24:6}=\frac{-3}{-4}\)

\(\frac{-35}{-70}=\frac{-35:35}{-70:35}=\frac{-1}{-2}\)

\(\frac{-9}{-27}=\frac{-9:9}{-27:9}=\frac{-1}{-3}\)

a) \(\frac{-9}{12}=\frac{-9\div3}{12\div3}=\frac{-3}{4}\)

\(\frac{-18}{-24}=\frac{-18\div\left(-6\right)}{-24\div\left(-6\right)}=\frac{3}{4}\)

\(\frac{-35}{70}=\frac{-35\div35}{70\div35}=\frac{-1}{2}\)

\(\frac{-9}{27}=\frac{-9\div9}{27\div9}=\frac{-1}{3}\)

b) \(\frac{1313}{4242}=\frac{1313\div101}{4242\div101}=\frac{13}{42}\)

\(\frac{-353535}{-424242}=\frac{-353535\div\left(-70707\right)}{-424242\div\left(-70707\right)}=\frac{5}{6}\)

c)\(\frac{2^3\times4^3\times5^4}{8^2\times25^3\times7}\)= \(\frac{2^3\times4^3\times5^4}{\left(2^3\right)^2\times\left(5^2\right)^3\times7}\)= \(\frac{2^3\times4^3\times5^4}{2^6\times5^6\times7}\)= \(\frac{2^3\times4^3\times5^4}{2^3\times2^3\times5^2\times5^4\times7}\)= \(\frac{2^3\times2^3}{2^3\times5^2\times7}\)= \(\frac{8}{25\times7}\)= \(\frac{8}{175}\)

talaays đơn thức nhân với từng hạng tử của đa thức

rồi cộng tích lại với nhau

rồi tìm x

nha bn

bạn giải luôn giúp mình được không ạ?

tìm x

a,(3x-1)(2x+7)-(x+1)(6x-5)=16

b, (10x+9)x-(5x-1)(2x+3)=8

c, (3x-5)(7-5x)+(5x+2)(3x-2)-2=0

d, x(x+1)(x+6)-x³=5x

Giúp mình vs ạ, mình cám ơn trước nha, mình đang cần gấp lắm ạ

\(x+x\cdot3:\dfrac{2}{9}+x:\dfrac{2}{7}=252\)

\(\Leftrightarrow x+x\cdot3\cdot\dfrac{9}{2}+x\cdot\dfrac{7}{2}=252\)

\(\Leftrightarrow x\cdot18=252\)

hay x=14