a: =>x-2/5=3/4:1/3=3/4*3=9/4

=>x=9/4+2/5=45/20+8/20=53/20

b: =>x-2/3=7/3:4/5=7/3*5/4=35/12

=>x=35/12+2/3=43/12

c: 1/3(x-2/5)=4/5

=>x-2/5=4/5*3=12/5

=>x=12/5+2/5=14/5

d: =>2/3x-1/3-1/4x+1/10=7/3

=>5/12x-7/30=7/3

=>5/12x=7/3+7/30=77/30

=>x=77/30:5/12=154/25

e: \(\Leftrightarrow x\cdot\dfrac{3}{7}-\dfrac{2}{7}+\dfrac{1}{2}-\dfrac{5}{4}x+\dfrac{5}{2}=0\)

=>\(x\cdot\dfrac{-23}{28}=\dfrac{2}{7}-3=\dfrac{-19}{7}\)

=>x=19/7:23/28=76/23

f: =>1/2x-3/2+1/3x-4/3+1/4x-5/4=1/5

=>13/12x=1/5+3/2+4/3+5/4=257/60

=>x=257/65

i: =>x^2-2/5x-x^2-2x+11/4=4/3

=>-12/5x=4/3-11/4=-17/12

=>x=17/12:12/5=85/144

a) \(\left(3x-1\right).\left(\frac{-1}{2}x+5\right)=0\)

\(\Rightarrow3x-1=0\Rightarrow3x=1\Rightarrow x=\frac{1}{3}\)

\(\frac{-1}{2}x+5=0\Rightarrow\frac{-1}{2}x=-5\Rightarrow x=10\)

b) \(3\left(x-\frac{1}{2}\right)-5\left(x+\frac{3}{5}\right)=x+\frac{1}{5}\)

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

\(\Rightarrow3x-5x-x=\frac{1}{5}+\frac{3}{2}+3\)

\(-3x=\frac{47}{10}\)

\(x=\frac{-47}{30}\)

c) \(-5.\left(x+\frac{1}{5}\right)-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{3}{2}x-\frac{5}{6}\)

\(-5x-1-\frac{1}{2}x+\frac{1}{3}=\frac{3}{2}x-\frac{5}{6}\)

\(-5x-\frac{1}{2}x-\frac{3}{2}x=\frac{-5}{6}+1-\frac{1}{3}\)

\(-7x=\frac{-1}{6}\)

\(x=\frac{1}{42}\)

d) \(3.\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)

\(3.\left(3x-\frac{1}{2}\right)^3=\frac{-1}{9}\)

\(\left(3x-\frac{1}{2}\right)^3=\frac{-1}{27}\)

\(\left(3x-\frac{1}{2}\right)^3=\left(\frac{-1}{3}\right)^3\)

\(\Rightarrow3x-\frac{1}{2}=\frac{-1}{3}\)

\(3x=\frac{1}{6}\)

\(x=\frac{1}{18}\)

Học tốt nhé bn!
 

x = \(\frac{1}{18}\)nha

x=1/8

 HT!

1) \(2^3\times x-5^2\times x=2\times\left(5^2+2^2\right)-33\)

\(x\times\left(2^3-5^2\right)=2\times\left(25+4\right)-33\)

\(x\times\left(8-25\right)=2\times29-33\)

\(x\times-17=25\)

\(x=-\dfrac{25}{17}\)

2) \(15\div\left(x+2\right)=\left(3^3+3\right)\div1\)

\(15\div\left(x+2\right)=\left(27+3\right)\div1\)

\(15\div\left(x+2\right)=30\div1\)

\(15\div\left(x+2\right)=30\)

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

\(x=-\dfrac{3}{2}\)

3) \(20\div\left(x+1\right)=\left(5^2+1\right)\div13\)

\(20\div\left(x+1\right)=\left(25+1\right)\div13\)

\(20\div\left(x+1\right)=26\div13\)

\(20\div\left(x+1\right)=2\)

\(x+1=20\div2\)

\(x+1=10\)

\(x=9\)

4) \(320\div\left(x-1\right)=\left(5^3-5^2\right)\div4+15\)

\(320\div\left(x-1\right)=\left(125-25\right)\div4+15\)

\(320\div\left(x-1\right)=100\div4+15\)

\(320\div\left(x-1\right)=25+15\)

\(320\div\left(x-1\right)=40\)

\(x-1=8\)

\(x=9\)

5) \(240\div\left(x-5\right)=2^2\times5^2-20\)

\(240\div\left(x-5\right)=4\times25-20\)

\(240\div\left(x-5\right)=100-20\)

\(240\div\left(x-5\right)=80\)

\(x-5=30\)

\(x=35\)

6) \(70\div\left(x-3\right)=\left(3^4-1\right)\div4-10\)

\(70\div\left(x-3\right)=\left(81-1\right)\div4-10\)

\(70\div\left(x-3\right)=80\div4-10\)

\(70\div\left(x-3\right)=20-10\)

\(70\div\left(x-3\right)=10\)

\(x-3=7\)

\(x=10\)

Bạn cần viết đề bài bằng công thức toán để được hỗ trợ tốt hơn. 

x^2+2x-3/3+2x/4=x^2/3

1.

$(x-2)(x-5)=(x-3)(x-4)$

$\Leftrightarrow x^2-7x+10=x^2-7x+12$
$\Leftrightarrow 10=12$ (vô lý)

Vậy pt vô nghiệm.

2.

$(x-7)(x+7)+x^2-2=2(x^2+5)$

$\Leftrightarrow x^2-49+x^2-2=2x^2+10$
$\Leftrightarrow 2x^2-51=2x^2+10$

$\Leftrightarrow -51=10$ (vô lý)

Vậy pt vô nghiệm.

3.

$(x-1)^2+(x+3)^2=2(x-2)(x+2)$
$\Leftrightarrow (x^2-2x+1)+(x^2+6x+9)=2(x^2-4)$
$\Leftrightarrow 2x^2+4x+10=2x^2-8$

$\Leftrightarrow 4x+10=-8$

$\Leftrightarrow 4x=-18$

$\Leftrightarrow x=-4,5$

4.

$(x+1)^2=(x+3)(x-2)$

$\Leftrightarrow x^2+2x+1=x^2+x-6$

$\Leftrightarrow x=-7$ 

5.

$x^2-(2x-1)(x+3)=3-x(5+x)$
$\Leftrightarrow x^2-(2x^2+5x-3)=3-(5x+x^2)$
$\Leftrightarro -x^2-5x+3=3-5x-x^2$ (luôn đúng)

Vậy pt có nghiệm $x\in\mathbb{R}$

6.

$3(5-2x)-4(x+2)=5x-18$

$\Leftrightarrow 15-6x-4x-8=5x-18$

$\Leftrightarrow 7-10x=5x-18$

$\Leftrightarrow 25=15x$

$\Leftrightarrow x=\frac{5}{3}$

Dễ thế mà không làm được thì bạn nên xem lại nhé,một hai câu thì còn được chứ cả 10 câu thế kia rõ là ỷ lại rồi bạn ạ.Thân!

ừ đúng dễ mà

1: Ta có: \(x^2-2x+5-\left(x-7\right)\left(x+2\right)\)

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

\(=3x-9\)

2: Ta có: \(-5x\left(x-5\right)+\left(x-3\right)\left(x^2-7\right)\)

\(=-5x^2+25x+x^3-7x-3x^2+21\)

\(=x^3-8x^2+18x+21\)

3: Ta có: \(x\left(x^2-x-2\right)-\left(x+5\right)\left(x-1\right)\)

\(=x^3-x^2-2x-x^2-4x+5\)

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

1) \(\left|x+\frac{4}{5}\right|+\frac{7}{5}=\frac{3}{5}\)

\(\Rightarrow\left|x+\frac{4}{5}\right|=\frac{3}{5}-\frac{7}{5}\)

\(\Rightarrow\left|x+\frac{4}{5}\right|=\frac{-4}{5}\)

\(x+\frac{4}{5}=\pm\frac{4}{5}\)

\(TH1:x+\frac{4}{5}=\frac{4}{5}\)

\(\Rightarrow x=\frac{4}{5}-\frac{4}{5}=0\)

\(TH2:x+\frac{4}{5}=\frac{-4}{5}\)

\(\Rightarrow x=\frac{-4}{5}-\frac{4}{5}=\frac{-8}{5}\)

Vậy x ∈ {0; \(\frac{-8}{5}\)}

Hai câu cuối khó nhìn nên ko giải

5,\(\left(x-\frac{1}{3}\right)^3=-\frac{1}{27}\)

Có \(-\frac{1}{27}=\left(-\frac{1}{3}\right)^3\)

\(\Rightarrow\left(x-\frac{1}{3}\right)^3=\left(-\frac{1}{3}\right)^3\)

TH1: \(\left(x-\frac{1}{3}\right)^3=\left(-\frac{1}{3}\right)^3\)

\(\Rightarrow x-\frac{1}{3}=-\frac{1}{3}\)

\(x=-\frac{1}{3}+\frac{1}{3}\Rightarrow x=0\)

Vậy x = 0