\(\frac{3x-7}{5}=\frac{2x-1}{3}\)

\(\Leftrightarrow9x-21=10x-5\)

\(\Leftrightarrow-x=16\Leftrightarrow x=-16\)

\(\frac{4x-7}{12}-x=\frac{3x}{8}\)

\(\Leftrightarrow\frac{4x-7-12x}{12}=\frac{3x}{8}\)

\(\Leftrightarrow\frac{-7-8x}{12}=\frac{3x}{8}\)

\(\Leftrightarrow-56-64x=36x\)

\(\Leftrightarrow-56=100x\Leftrightarrow x=\frac{-14}{25}\)

\(\frac{x-2009}{1234}+\frac{x-2009}{5678}-\frac{x-2009}{197}=0\)

\(\Leftrightarrow\left(x-2019\right)\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)=0\)

Vì \(\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)\ne0\)nên x - 2019 = 0

Vậy x = 2019

\(\frac{5x-8}{3}=\frac{1-3x}{2}\)

\(\Leftrightarrow10x-16=3-9x\)

\(\Leftrightarrow19x=19\Leftrightarrow x=1\)

\(\frac{x-5}{6}-\frac{x-9}{4}=\frac{5x-3}{8}+2\)

\(\Rightarrow\frac{4x-20-6x+54}{24}=\frac{5x-3+16}{8}\)

\(\Rightarrow\frac{-2x+34}{24}=\frac{5x+13}{8}\)

\(\Rightarrow-16x-272=120x+312\)

\(\Leftrightarrow-136x=584\Leftrightarrow x=\frac{-73}{17}\)

a, làm tương tự với phần b bài nãy bạn đăng 

b, \(\left(x+1\right)^2-5=x^2+11\)

\(\Leftrightarrow x^2+2x+1-5=x^2+11\)

\(\Leftrightarrow2x-10=0\Leftrightarrow x=5\)

Vậy tập nghiệm của phương trình là S = { 5 } ( kết luận như thế với các phần sau nhé ! ) 

c, \(3\left(3x-1\right)=3x+5\Leftrightarrow9x-3-3x-5=0\)

\(\Leftrightarrow6x-8=0\Leftrightarrow x=\frac{4}{3}\)

d, \(3x\left(2x-3\right)-3\left(3+2x^2\right)=0\)

\(\Leftrightarrow6x^2-9x-9-6x^2=0\Leftrightarrow-9x=9\Leftrightarrow x=-1\)

e, khai triển nó ra rút gọn rồi giải thôi nhé! ( tự làm )

f, \(\left(x-1\right)^2-x\left(x+1\right)+3\left(x-2\right)+5=0\)

\(\Leftrightarrow x^2-2x+1-x^2+x+3x-6+5=0\)

\(\Leftrightarrow2x=0\Leftrightarrow x=\frac{0}{2}\)vô lí 

Vậy phương trình vô nghiệm 

a,  \((\frac{3}{7}-\frac{2}{3})\)  .x  =\(\frac{10}{21}\)                                                                   

  \(\frac{-5}{21}\).x=\(\frac{10}{21}\)

x= -2

 Mk chỉ làm 1 phần  các phằn còn lại tương tự

Bài rút gọn biểu thức

a) M=|2x-3|+|x-1| với x > 1,5

b) N=|2-x|-3|x+1| với x < -1

c) P=|3x-5|+|x-2|

d) Q=|x-3|-2.|-5x|

a, \(\frac{3}{7}x-\frac{2}{3}x=\frac{10}{21}\Rightarrow\left(\frac{3}{7}-\frac{2}{3}\right)x=\frac{10}{21}\Rightarrow x=\frac{10}{21}.\left(\frac{-21}{5}\right)\Rightarrow x=-2\)

b, \(\frac{7}{35}:\left(x-\frac{1}{3}\right)=\frac{2}{25}\Rightarrow x-\frac{1}{3}=\frac{1}{5}:\frac{2}{25}\Rightarrow x=\frac{1}{10}-\frac{1}{3}=\frac{13}{30}\)

c, \(|2x-4|+1=5\)

\(\Rightarrow\orbr{\begin{cases}2x-4=4\\2x-4=-4\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=0\end{cases}}\)

d, \(3.|3-2x|-1=\frac{2}{5}\)

\(\Rightarrow3.|3-2x|=\frac{7}{5}\)

\(\Rightarrow\orbr{\begin{cases}3-2x=\frac{7}{15}\\3-2x=-\frac{7}{15}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{19}{15}\\x=\frac{26}{15}\end{cases}}\)

f, \(\left(2x-1\right).\left(x+\frac{2}{3}\right)=0\)

\(\Rightarrow\hept{\begin{cases}2x-1=0\\x+\frac{2}{3}=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\x=-\frac{2}{3}\end{cases}}\)

a) \(\left|x-1\right|+\left|x+3\right|=4\left(1\right)\)

+) TH1: Nếu \(x< -3\) thì \(x-1< 0;x+3< 0\)

\(\Rightarrow\left|x-1\right|=-x+1;\left|x+3\right|=-x-3\)

PT (1) trở thành: \(-x+1-x-3=4\)

\(\Leftrightarrow-2x=6\Leftrightarrow x=-3\left(loại\right)\)

+) TH2: Nếu \(-3\le x< 1\) thì \(x-1< 0;x+3>0\)

\(\Rightarrow\left|x-1\right|=-x+1;\left|x+3\right|=x+3\)

PT (1) trở thành: \(-x+1+x+3=4\)

\(\Leftrightarrow0x=0\) (luôn đúng)

Kết hợp với đk ta được: \(\Rightarrow-3\le x< 1\)

+) TH3: Nếu \(x\ge1\) thì \(x-1>0;x+3>0\)

\(\Rightarrow\left|x-1\right|=x-1;\left|x+3\right|=x+3\)

PT (1) trở thành: \(x-1+x+3=4\)

\(\Leftrightarrow2x=2\Leftrightarrow x=1\left(t/m\right)\)

Vậy x nằm trong khoảng \(-3\le x\le1.\)

Mấy bài kia làm tương tự.

2.

\(\left|x+1\right|+\left|x+2\right|+...+\left|x+10\right|=605x\)(1)

Vì các thừa số ở vế phải của (1) đều không âm nên x không âm. Do đó \(\left|x+1\right|+\left|x+2\right|+...+\left|x+10\right|=\left(x+1\right)+\left(x+2\right)+...+\left(x+10\right)\)

\(\Rightarrow\left(x+1\right)+\left(x+2\right)+...+\left(x+10\right)=605x\)

\(\Rightarrow10x+\dfrac{10\left(10+1\right)}{2}=605x\)

\(\Rightarrow55=595x\)

\(\Rightarrow x=\dfrac{55}{595}=\dfrac{11}{119}\)

Vậy x = \(\dfrac{11}{119}\)

2.

a) Ta có \(\left|3x-2019\right|=2019-3x\)

khi \(3x-2019< 0\)

\(\Leftrightarrow x< \dfrac{2019}{3}=673\)

Vậy \(x< 673.\)

b) Vì \(\left|x+1\right|\ge0\forall x;...\left|x+10\right|\ge0\forall x\)

\(\Rightarrow\left|x+1\right|+...+\left|x+10\right|\ge0\forall x\)

nên \(605x\ge0\Leftrightarrow x\ge0\)

\(\Rightarrow\left|x+1\right|=x+1;...\left|x+10\right|=x+10\)

Ta có: \(x+1+...+x+10=605x\)

\(\Rightarrow10x+55=605x\Rightarrow x=\dfrac{11}{119}\)

Vậy \(x=\dfrac{11}{119}.\)

e, 3x(2-x) =15(x-2)

\(\Leftrightarrow3x\left(2-x\right)-15\left(x-2\right)=0\)

\(\Leftrightarrow-3x\left(x-2\right)-15\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(-3x-15\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\-3x-15=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)

Vậy..

f, (x+5)(x+4)=0

\(\Leftrightarrow\left\{{}\begin{matrix}x+5=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\x=-4\end{matrix}\right.\)

Vậy..

g, x(x+4)=0

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)

,h, (2x -4)(x-2)=0

\(\Leftrightarrow2\left(x-2\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(2-1\right)=0\)

\(\Leftrightarrow x-2=0\Leftrightarrow x=2\)

i, (x+1/5)(2x-3)=0

\(\Leftrightarrow\left\{{}\begin{matrix}x+\frac{1}{5}=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{-1}{5}\\x=\frac{3}{2}\end{matrix}\right.\)

k, x²-4x=0

\(\Leftrightarrow x\left(x-2\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

m, 4x²-1=0

\(\Leftrightarrow\left(2x\right)^2-1^2=0\)

\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)=0\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{2}\\x=\frac{-1}{2}\end{matrix}\right.\)

n, x²-6x+9=0

\(\Leftrightarrow x^2-2.x.3+3^2=0\)

\(\Leftrightarrow\left(x-3\right)^2=0\Leftrightarrow x-3=0\)

<=> x=3

l, (3x-5)²-(x+4)²=0

\(\Leftrightarrow\left(3x-5-x-4\right)\left(3x-5+x+4\right)=0\)

\(\Leftrightarrow\left(2x-9\right)\left(4x-1\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x-9=0\\4x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=9\\4x=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{9}{2}\\x=\frac{1}{4}\end{matrix}\right.\)

Vậy ..

o, 7x(x+2)-5(x+2)=0

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

\(\Leftrightarrow\left\{{}\begin{matrix}x+2=0\\7x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\7x=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x=\frac{5}{7}\end{matrix}\right.\)

Vậy....

p, 3x(2x-5)-4x+10=0

\(\Leftrightarrow3x\left(2x-5\right)-\left(4x-10\right)=0\)

\(\Leftrightarrow3x\left(2x-5\right)-2\left(2x-5\right)=0\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}2x-5=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2x=5\\3x=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{5}{2}\\x=\frac{2}{3}\end{matrix}\right.\)

Vậy...

q, (2-2x)-x²+1=0

\(\Leftrightarrow2\left(1-x\right)-\left(x^2-1^2\right)=0\)

\(\Leftrightarrow2\left(1-x\right)-\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow2\left(1-x\right)+\left(1-x\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(1-x\right)\left(2+x+1\right)=0\)

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

Vậy ....

r, x(1-3x)=5(1-3x)

\(\Leftrightarrow x\left(1-3x\right)-5\left(1-3x\right)=0\)

\(\Leftrightarrow\left(1-3x\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}1-3x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-3x=-1\\x=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{3}\\x=5\end{matrix}\right.\)

s, 2x-3/4+x+1/6=3

\(\Leftrightarrow x-\frac{7}{12}=3\Leftrightarrow x=3+\frac{7}{12}=\frac{43}{12}\)

r, x(1-3x)=5(1-3x)

➜x(1-3x)-5(1-3x)=0

➜(x-5)(1-3x)=0

➜\(\left[{}\begin{matrix}x-5=0\\1-3x=0\end{matrix}\right.\text{➜}\left[{}\begin{matrix}x=5\\x=\frac{1}{3}\end{matrix}\right.\)

Mk lười lắm mai nha!!!~~~~~~~~~~~~

Làm dần:

a, 3x+6=0

➜3x=-6

➜x=2

b, 2x-10=0

➜2x=10

➜x=5

c, 3x-7=11

➜3x=11+7

➜3x=18

➜x=6

d, 3x-9=0

➜3x=9

➜x=3

\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)

\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)

Bài 4:

a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)

\(\Leftrightarrow6x-9-2x+4=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

hay \(x=\dfrac{13}{3}\)

c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

hay x=1

a/ \(3\left(2x-3\right)+2\left(2-x\right)=-3\)

\(\Leftrightarrow6x-9+4-2x=-3\)

\(\Leftrightarrow4x=2\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

Vậy: \(x=\dfrac{1}{2}\)

===========

b/ \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

\(\Leftrightarrow x=\dfrac{13}{3}\)

Vậy: \(x=\dfrac{13}{3}\)

==========

c/  \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

d/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)

\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=8\)

\(\Leftrightarrow-2x=-2\)

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

e/ \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x-16+11\)

\(\Leftrightarrow-14x=-4\)

\(\Leftrightarrow x=\dfrac{2}{7}\)

Vậy: \(x=\dfrac{2}{7}\)

==========

f/ \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)

\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)

\(\Leftrightarrow-x^3=8\)

\(\Leftrightarrow x=-2\)

Vậy: \(x=-2\)

a: \(x\in\left\{0;25\right\}\)

c: \(x\in\left\{0;5\right\}\)

Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^