a) \(\dfrac{3x-4}{2x+5}=\dfrac{3x+7}{2x-20}\left(đk:x\ne-\dfrac{5}{2},x\ne10\right)\)

\(\Rightarrow\left(3x-4\right)\left(2x-20\right)=\left(3x+7\right)\left(2x+5\right)\)

\(\Rightarrow6x^2-68x+80=6x^2+29x+35\)

\(\Rightarrow97x=45\Rightarrow x=\dfrac{45}{97}\)

b) \(\dfrac{10x-5}{7x+2}=\dfrac{50x+10}{35x-29}\left(đk:x\ne-\dfrac{2}{7},x\ne\dfrac{29}{35}\right)\)

\(\Rightarrow\left(10x-5\right)\left(35x-29\right)=\left(50x+10\right)\left(7x+2\right)\)

\(\Rightarrow350x^2-465x+145=350x^2+170x+20\)

\(\Rightarrow635x=125\Rightarrow x=\dfrac{25}{127}\)

b) | -5 | . (-7) + 4 . (-9)

= 5 . (-7) + (-36)

= -35 + (-36) = -71

\(a,0,5x-\frac{2}{3}x=\frac{7}{12}\Rightarrow\frac{1}{2}x-\frac{2}{3}x=\frac{7}{12}\)

\(\Rightarrow x\left(\frac{1}{2}-\frac{2}{3}\right)=\frac{7}{12}\Rightarrow x\cdot\left(\frac{3}{6}-\frac{4}{6}\right)=\frac{7}{12}\)

\(\Rightarrow x\cdot\left(-1\right)=\frac{7}{12}\Rightarrow x=\frac{7}{12}:\left(-1\right)=\frac{7}{-12}\)

\(c,\frac{\left(x-5\right)}{12}\cdot\frac{9}{29}=\frac{-6}{29}\Rightarrow\frac{\left(x-5\right)}{12}=\frac{-6}{29}:\frac{9}{26}\)

\(\frac{\Rightarrow\left(x-5\right)}{12}=\frac{-6}{9}=\frac{-2}{3}\Rightarrow x-5=-\frac{2}{3}\cdot12\)

\(\Rightarrow x-5=\frac{-24}{3}=-8\Rightarrow x=-8+5=-3\)

\(a,0,5x-\frac{2}{3}x=\frac{7}{12}\)

\(\Rightarrow\frac{1}{2}x-\frac{2}{3}x=\frac{7}{12}\)

\(\Rightarrow-\frac{1}{6}x=\frac{7}{12}\)

\(\Rightarrow x=-\frac{7}{2}\)

\(c,\frac{x-5}{12}\cdot\frac{9}{29}=-\frac{6}{29}\)

\(\Rightarrow\frac{x-5}{12}=-\frac{2}{3}\)

\(\Rightarrow x-5=12.\left(-\frac{2}{3}\right)\)

\(\Rightarrow x-5=-8\)

\(\Rightarrow x=-3\)

b)\(\frac{-8}{17}+\frac{5}{17}< \frac{x}{17}< \frac{-6}{17}+\frac{9}{17}\)\(\left(x\in Z\right)\)

\(\Rightarrow\frac{-3}{17}< \frac{x}{17}< \frac{3}{17}\)

\(\Rightarrow-3< x< 3\)

\(\Rightarrow x\in\left\{-2;-1;0;1;2\right\}\)

Vậy \(x\in\left\{-2;-1;0;1;2\right\}\)

a) (x-17)/33 + (x-21)/29 + x/25 = 4 <=> [(x-17)/33-1] + [(x-21)/29-1] + x/25-2] = 0 <=> (x-50)/33 + (x-50)/29 + (x-50)/25 = 0 <=> (x-50)(1/33+1/29+1/25) = 0 Mà 1/33+1/29+1/25 khác 0. <=> x- 50 = 0 <=> x=50 b) (3x−5)(7−5x)+(5x+2)(3x−2)=2 <=> 21x−15x2−35+25x+15x2−10x+6x−4−2=0 <=> (15x2−15x2)+(25x+21x−10x+6x)−(35+4+2)=0 <=> 42x=41 <=> x=41/42

Cộng từng hạng tử của hai vế với 1 , ta suy ra được tử chung là 3x+21=0

=> x=-7

c.

ĐLXĐ: \(x\ge-\dfrac{1}{3}\)

\(-\left(3x+1\right)+\sqrt{3x+1}+4x^2-10x+6=0\)

Đặt \(\sqrt{3x+1}=t\ge0\)

\(\Rightarrow-t^2+t+4x^2-10x+6=0\)

\(\Delta=1+4\left(4x^2-10x+6\right)=\left(4x-5\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{-1+4x-5}{-2}=3-2x\\t=\dfrac{-1-4x+5}{-2}=2x-2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3x+1}=3-2x\left(x\le\dfrac{3}{2}\right)\\\sqrt{3x-1}=2x-2\left(x\ge1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+1=4x^2-12x+9\left(x\le\dfrac{3}{2}\right)\\3x-1=4x^2-8x+4\left(x\ge1\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)

b.

ĐKXĐ: \(x\ge-\dfrac{61}{12}\)

\(\Leftrightarrow36x^2+12x-58-2\sqrt{12x+61}=0\)

\(\Leftrightarrow\left(36x^2+24x+4\right)-\left(12x+61+2\sqrt{12x+61}+1\right)=0\)

\(\Leftrightarrow\left(6x+2\right)^2-\left(\sqrt{12x+61}+1\right)^2=0\)

\(\Leftrightarrow\left(6x+1-\sqrt{12x+61}\right)\left(6x+3+\sqrt{12x+61}\right)=0\)

\(\Leftrightarrow...\) tương tự câu a

a.

ĐKXĐ: \(x\ge-\dfrac{5}{4}\)

\(\Leftrightarrow4x^2-12x-2-2\sqrt{4x+5}=0\)

\(\Leftrightarrow\left(4x^2-8x+4\right)-\left(4x+5+2\sqrt{4x+5}+1\right)=0\)

\(\Leftrightarrow\left(2x-2\right)^2-\left(\sqrt{4x+5}+1\right)^2=0\)

\(\Leftrightarrow\left(2x-2-\sqrt{4x+5}-1\right)\left(2x-2+\sqrt{4x+5}+1\right)=0\)

\(\Leftrightarrow\left(2x-3-\sqrt{4x+5}\right)\left(2x-1+\sqrt{4x+5}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{4x+5}=2x-3\left(x\ge\dfrac{3}{2}\right)\\\sqrt{4x+5}=1-2x\left(x\le\dfrac{1}{2}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+5=4x^2-12x+9\left(x\ge\dfrac{3}{2}\right)\\4x+5=4x^2-4x+1\left(x\le\dfrac{1}{2}\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)

a) \(2x\left(x-5\right)-x\left(3+2x\right)=26\)

\(\Rightarrow2x^2-10x-3x-2x^2=26\)

\(\Rightarrow-13x=26\Rightarrow x=-2\)

b) \(3x\left(1-2x\right)+2\left(3x+7\right)=29\)

\(\Rightarrow3x-6x^2+6x+14=29\)

\(\Rightarrow-6x^2+9x-15=0\)

\(\Rightarrow-6\left(x^2-\dfrac{3}{2}x+\dfrac{9}{16}\right)-\dfrac{93}{8}=0\)

\(\Rightarrow-6\left(x-\dfrac{3}{4}\right)^2-\dfrac{93}{8}=0\)(vô lý)

Vậy \(S=\varnothing\)

a. \(2x^2-10x-3x-2x^2=26\Leftrightarrow-13x=26\Leftrightarrow x=-2\)

a: \(\Leftrightarrow2x^2-10x-3x-2x^2=26\)

hay x=-2

\(C=x^3+3x+3x^2+5\)

\(=\left(x^3+3x^2+3x+1\right)+4\)

\(=\left(x+1\right)^3+4\)

Thay \(x=29\) vào biểu thức C ta được :

\(C=\left(29+1\right)^3+4=30^3+4=27000+4=27004\)

Vậy................

\(D=x^3-3x^2+3x\)

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

\(=\left(x-1\right)^3+1\)

Thay \(x=11\) vào biểu thức D ta được :

\(D=\left(11-1\right)^3+1=10^3+1=1000+1=1001\)

Vậy..................

Thay \(x=29\) vào C , ta được:

\(C=29^3+3.29+3.29^2+5\)

\(C=24389+87+2523+5\)

\(C=27004\)

Thay \(x=11\) vào D, ta được:

C=\(11^3-3.11^2+3.11\)

\(C=1331-363+33\)

\(C=1001\)

Thay x = 29 vào C ta có :

C = 29³ + 3 . 29 + 3 . 29² + 5

C = 24839 + 87 + 2523 + 5

C = 270004

Thay x = 11 vào D ta có :

D = 11³ - 3 .11² + 3 . 11

D = 1331 - 363 + 33

D = 1001