`(3x+2)/3 <= (x-4)/7`

`<=>7(3x+2) <= 3(x-4)`

`<=> 21x+14<=3x-12`

`<=>18x <= -26`

`<=> x <=-13/9`

\(a,3-x=x+1,8\)

\(\Rightarrow-x-x=1,8-3\)

\(\Rightarrow-2x=-1,2\)

\(\Rightarrow x=0,6\)

\(b,2x-5=7x+35\)

\(\Rightarrow2x-7x=35+5\)

\(\Rightarrow-5x=40\)

\(\Rightarrow x=-8\)

\(c,2\left(x+10\right)=3\left(x-6\right)\)

\(\Rightarrow2x+20=3x-18\)

\(\Rightarrow2x-3x=-18-20\)

\(\Rightarrow-x=-38\)

\(\Rightarrow x=38\)

\(d,8\left(x-\dfrac{3}{8}\right)+1=6\left(\dfrac{1}{6}+x\right)+x\)

\(\Rightarrow8x-3+1=1+6x+x\)

\(\Rightarrow8x-3=7x\)

\(\Rightarrow8x-7x=3\)

\(\Rightarrow x=3\)

\(e,\dfrac{2}{9}-3x=\dfrac{4}{3}-x\)

\(\Rightarrow-3x+x=\dfrac{4}{3}-\dfrac{2}{9}\)

\(\Rightarrow-2x=\dfrac{10}{9}\)

\(\Rightarrow x=-\dfrac{5}{9}\)

\(g,\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{3}{4}x-\dfrac{1}{2}\)

\(\Rightarrow\dfrac{1}{2}x-\dfrac{3}{4}x=-\dfrac{1}{2}-\dfrac{5}{6}\)

\(\Rightarrow-\dfrac{1}{4}x=-\dfrac{4}{3}\)

\(\Rightarrow x=\dfrac{16}{3}\)

\(h,x-4=\dfrac{5}{6}\left(6-\dfrac{6}{5}x\right)\)

\(\Rightarrow x-4=5-x\)

\(\Rightarrow x+x=5+4\)

\(\Rightarrow2x=9\)

\(\Rightarrow x=\dfrac{9}{2}\)

\(k,7x^2-11=6x^2-2\)

\(\Rightarrow7x^2-6x^2=-2+11\)

\(\Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

\(m,5\left(x+3\cdot2^3\right)=10^2\)

\(\Rightarrow5\left(x+24\right)=100\)

\(\Rightarrow x+24=20\)

\(\Rightarrow x=-4\)

\(n,\dfrac{4}{9}-\left(\dfrac{1}{6^2}\right)=\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}\)

\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{4}{9}-\dfrac{1}{36}\)

\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{5}{12}\)

\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2=0\)

\(\Rightarrow x-\dfrac{2}{3}=0\Rightarrow x=\dfrac{2}{3}\)

#\(Urushi\text{☕}\)

a: 3-x=x+1,8

=>-2x=-1,2

=>x=0,6

b: 2x-5=7x+35

=>-5x=40

=>x=-8

c: 2(x+10)=3(x-6)

=>3x-18=2x+20

=>x=38

d; 8(x-3/8)+1=6(1/6+x)+x

=>8x-3+1=1+6x+x

=>8x-2=7x+1

=>x=3

e: =>-3x+x=4/3-2/9

=>-2x=12/9-2/9=10/9

=>x=-5/9

g: =>3/4x-1/2x=5/6+1/2

=>1/4x=5/6+3/6=8/6=4/3

=>x=4/3*4=16/3

h: =>x-4=-x+5

=>2x=9

=>x=9/2

bạn tách từng bài ra bn

tl câu hỏi trên cho mik ik

a, \(\dfrac{t+3}{t-2}+\dfrac{t-2}{t+3}=\dfrac{5t+15}{t^2+t-6}\) ĐKXĐ: t\(\ne\)2,t\(\ne\)-3

\(\Leftrightarrow\dfrac{t+3}{t-2}+\dfrac{t-2}{t+3}=\dfrac{5t+15}{\left(t+3\right)\left(t-2\right)}\)

\(\Rightarrow\left(t+3\right)\left(t+3\right)+\left(t-2\right)\left(t-2\right)=5t+15\)

\(\Leftrightarrow t^2+6t+9+t^2-4t+4-5t-15=0\)

\(\Leftrightarrow-3t-2=0\)

\(\Leftrightarrow-3t=2\)

\(\Leftrightarrow t=\dfrac{-2}{3}\) (tđk)

\(\Rightarrow S=\left\{\dfrac{-2}{3}\right\}\)

b, \(\left(2x+3\right)\left(\dfrac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)\)ĐKXĐ: x\(\ne\)\(\dfrac{2}{7}\)

\(\Leftrightarrow\) \(\left(2x+3\right)\left(\dfrac{3x+8}{2-7x}+1\right)-\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)=0\)

\(\Rightarrow\left(\dfrac{3x+8}{2-7x}+1\right)\left(2x+3-x+5\right)=0\)

\(\Leftrightarrow\) \(\Rightarrow\left(\dfrac{3x+8}{2-7x}+1\right)\left(x+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{3x+8}{2-7x}+1=0\\x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x+8+2-7x=0\\x=-8\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-4x+10=0\\x=-8\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-8\end{matrix}\right.\)

\(\Rightarrow S=\left\{\dfrac{5}{2};-8\right\}\)

ĐKXĐ: x khác 2 và x khác -3

\(\dfrac{t+3}{t-2}+\dfrac{t-2}{t+3}=\dfrac{5t+15}{t^2+t-6}\)

\(\Leftrightarrow\dfrac{\left(t+3\right)\left(t+3\right)}{\left(t+3\right)\left(t-2\right)}+\dfrac{\left(t-2\right)\left(t-2\right)}{\left(t+3\right)\left(t-2\right)}=\dfrac{5t+15}{t^2+t-6}\)

\(\Rightarrow t^2+6t+9+t^2-4=5t+15\)

\(\Leftrightarrow2t^2+t-10=0\)

\(\Leftrightarrow2t^2-4t+5t-10=0\)

\(\Leftrightarrow2t\left(t-2\right)+5\left(t-2\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}t=2\left(loại\right)\\t=\dfrac{-5}{2}\end{matrix}\right.\)

Vậy..................

ĐKXĐ: x khác \(\dfrac{2}{7}\)\(\left(2x+3\right)\left(\dfrac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)\)

\(\Leftrightarrow\left(2x+3\right)\left(\dfrac{3x+8}{2-7x}+1\right)-\left(x+5\right)\left(\dfrac{3x+8}{2-7x}+1\right)=0\)

\(\Leftrightarrow\left(\dfrac{3x+8}{2-7x}+1\right)\left(2x+3-x-5\right)=0\)

\(\Leftrightarrow\left(\dfrac{-4x+10}{2-7x}\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{-4x+10}{2-7x}=0\\x-2=0\end{matrix}\right.\)

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

Vậy...............

1/ ĐKXĐ : \(x\ne1\)

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

\(\Leftrightarrow21x-9=2x-2\)

\(\Leftrightarrow19x=7\Leftrightarrow x=\dfrac{7}{19}\left(tm\right)\)

Vậy...

b/ \(\dfrac{2\left(3-7x\right)}{1+x}=\dfrac{1}{2}\) ĐKXĐ : \(x\ne-1\)

\(\Leftrightarrow12-28x=1+x\)

\(\Leftrightarrow11=29x\Leftrightarrow x=\dfrac{11}{29}\) \(\left(tm\right)\)

Vậy....

c/ ĐKXĐ : \(x\ne0\)

\(\dfrac{x^2-6}{x}=x+\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{x^2-6}{x}=\dfrac{2x+3}{2}\)

\(\Leftrightarrow2x^2-12=2x^2+3x\)

\(\Leftrightarrow3x=-12\Leftrightarrow x=-4\) \(\left(tm\right)\)

Vậy...

4/ ĐKXĐ : \(x\ne-\dfrac{2}{3}\)

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

\(\Leftrightarrow\left(2x-1\right)\left(3x+2\right)=5\)

\(\Leftrightarrow6x^2+4x-3x-2=5\)

\(\Leftrightarrow6x^2+x-7=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{6}\\x=1\end{matrix}\right.\)

Vậy....

5,6 Tương tự nhé !

1)ĐKXĐ: \(x\ne1\)

Ta có: \(\dfrac{7x-3}{x-1}=\dfrac{2}{3}\)

\(\Leftrightarrow3\left(7x-3\right)=2\left(x-1\right)\)

\(\Leftrightarrow21x-9=2x-2\)

\(\Leftrightarrow21x-9-2x+2=0\)

\(\Leftrightarrow19x-7=0\)

\(\Leftrightarrow19x=7\)

\(\Leftrightarrow x=\dfrac{7}{19}\)(nhận)

Vậy: \(S=\left\{\dfrac{7}{19}\right\}\)

2) ĐKXĐ: \(x\ne-1\)

Ta có: \(\dfrac{2\left(3-7x\right)}{1+x}=\dfrac{1}{2}\)

\(\Leftrightarrow4\left(3-7x\right)=x+1\)

\(\Leftrightarrow12-28x-x-1=0\)

\(\Leftrightarrow-29x+11=0\)

\(\Leftrightarrow-29x=-11\)

\(\Leftrightarrow x=\dfrac{11}{29}\)

Vậy: \(S=\left\{\dfrac{11}{29}\right\}\)

3) ĐKXĐ: \(x\ne0\)

Ta có: \(\dfrac{x^2-6}{x}=x+\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{x^2-6}{x}=\dfrac{2x+3}{2}\)

\(\Leftrightarrow2\left(x^2-6\right)=x\left(2x+3\right)\)

\(\Leftrightarrow2x^2-12=2x^2+6x\)

\(\Leftrightarrow2x^2-12-2x^2-6x=0\)

\(\Leftrightarrow-6x-12=0\)

\(\Leftrightarrow-6x=12\)

\(\Leftrightarrow x=-2\)

Vậy: S={-2}

1.

<=> \(\left[{}\begin{matrix}4-3x=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=2\end{matrix}\right.\)

2.

<=>\(\left[{}\begin{matrix}7-2x=0\\4+8x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)

3.

<=>\(\left[{}\begin{matrix}9-7x=0\\11-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{7}\\x=\dfrac{11}{3}\end{matrix}\right.\)

4.

<=>\(\left[{}\begin{matrix}7-14x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)

5. 

<=>\(\left[{}\begin{matrix}\dfrac{7}{8}-2x=0\\3x+\dfrac{1}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{16}\\x=-\dfrac{1}{9}\end{matrix}\right.\)

6,7. ko đủ điều kiện tìm

\(\dfrac{3-3x}{5}=\dfrac{x-1}{2}\\ \Leftrightarrow2\left(3-3x\right)=5x-5\\ \Leftrightarrow6-6x=5x-5\\ \Leftrightarrow-6x+5x=-5-6\\ \Leftrightarrow-x=-11\\ \Rightarrow x=11\)

1: \(=-3x^3-21x^2+x\)

2: \(=-15x^4y^7+10x^5y^6+5x^3y^5\)

3: \(=x^7y^4-2x^4y^2-3x^3y\)

5: \(=15x-6x^2\)

6: \(=4x^3-8x^2+10x\)

7: \(=-8x^5y^3+16x^7y^2-12x^3y^4\)

8: \(=x^7y^4-2x^4y^2-3x^3y\)