\(1,\dfrac{4x-3}{x-5}=\dfrac{29}{3}\left(ĐKXĐ:x\ne5\right)\)

\(\Rightarrow3\left(4x-3\right)=29\left(x-5\right)\)

\(\Leftrightarrow12x-9=29x-145\)

\(\Leftrightarrow12x-9-29x+145=0\)

\(\Leftrightarrow-17x+136=0\)

\(\Leftrightarrow-17x=-136\)

\(\Leftrightarrow x=8\left(tm\right)\)

Vậy \(S=\left\{8\right\}\)

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

\(\Rightarrow2x-1=2\left(5-3x\right)\)

\(\Leftrightarrow2x-1=10-6x\)

\(\Leftrightarrow2x-1-10+6x=0\)

\(\Leftrightarrow8x-11=0\)

\(\Leftrightarrow8x=11\)

\(\Leftrightarrow x=\dfrac{11}{8}\left(tm\right)\)

Vậy \(S=\left\{\dfrac{11}{8}\right\}\)

\(3,\dfrac{4x-5}{x-1}=2+\dfrac{x}{x-1}\left(ĐKXĐ:x\ne1\right)\)

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

\(\Leftrightarrow\dfrac{4x-5}{x-1}=\dfrac{2x-2}{x-1}+\dfrac{x}{x-1}\)

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

\(\Rightarrow4x-5=3x-2\)

\(\Leftrightarrow4x-5-3x+2=0\)

\(\Leftrightarrow x-3=0\)

\(\Leftrightarrow x=3\left(tm\right)\)

Vậy \(S=\left\{3\right\}\)

\(4,\dfrac{2x+5}{2x}-\dfrac{x}{x+5}=0\left(ĐKXĐ:x\ne\dfrac{1}{2};x\ne-5\right)\)

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

\(\Leftrightarrow\dfrac{2x^2+15x+25}{2x\left(x+5\right)}-\dfrac{2x^2}{2x\left(x+5\right)}=0\)

\(\Leftrightarrow\dfrac{15x+25}{2x\left(x+5\right)}=0\)

\(\Rightarrow15x+25=0\)

\(\Leftrightarrow15x=-25\)

\(\Leftrightarrow x=\dfrac{-5}{3}\left(tm\right)\)

Vậy \(S=\left\{\dfrac{-5}{3}\right\}\)

\(1,\dfrac{4x-3}{x-5}=\dfrac{29}{3}\)

\(\Leftrightarrow\dfrac{3\left(4x-3\right)-29\left(x-5\right)}{3\left(x-5\right)}=0\)

\(\Leftrightarrow12x-9-29x+145=0\)

\(\Leftrightarrow-17x=-136\)

\(\Leftrightarrow x=8\)

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

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

\(\Leftrightarrow2x-1-10+6x=0\)

\(\Leftrightarrow8x=11\)

\(\Leftrightarrow x=\dfrac{11}{8}\)

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

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

\(\Leftrightarrow4x-5-2x+2+2x=0\)

\(\Leftrightarrow4x=3\)

\(\Leftrightarrow x=\dfrac{3}{4}\)

\(4,\dfrac{2x+5}{2x}-\dfrac{x}{x+5}=0\)

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

\(\Leftrightarrow2x^2+10x+5x+25-2x^2=0\)

\(\Leftrightarrow15x=-25\)

\(\Leftrightarrow x=-\dfrac{5}{3}\)

\(\left|2x+7\right|-\dfrac{1}{2}=4\)

\(\Rightarrow\left|2x+7\right|=\dfrac{9}{2}\)

\(\Rightarrow\left[{}\begin{matrix}2x+7=\dfrac{9}{2}\\2x+7=-\dfrac{9}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=-\dfrac{5}{2}\\2x=-\dfrac{23}{2}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{4}\\x=-\dfrac{23}{4}\end{matrix}\right.\)

Vậy: \(x\in\left\{-\dfrac{5}{4};-\dfrac{23}{4}\right\}\).

\(\dfrac{-4}{x}=\dfrac{x}{-49}\\ \Rightarrow x^2=\left(-4\right)\left(-49\right)\\ \Rightarrow x^2=196\\ \Rightarrow x=\pm14\)

\(\dfrac{3.6}{x-3}=\dfrac{5}{3}\\ \Rightarrow5\left(x-3\right)=3.3.6\\ \Rightarrow5\left(x-3\right)=54\\ \Rightarrow x-3=\dfrac{54}{5}\\ \Rightarrow x=\dfrac{54}{5}+3\\ \Rightarrow x=\dfrac{69}{15}\)

\(\left(2x+1\right):2=12:3\\ \left(2x+1\right):2=4\\2x+1=2\\ 2x=1\\ x=\dfrac{1}{2} \)

\(\left(2x-14\right):3=12:9\\ \left(2x-14\right):3=\dfrac{4}{3}\\ 2x-14=4\\ 2x=16\\ x=8\)

Lời giải:

a.

 \(\frac{10}{x+2}=\frac{60}{6(x+2)}=\frac{60(x-2)}{6(x+2)(x-2)}=\frac{60(x-2)}{6(x^2-4)}\)

\(\frac{5}{2x-4}=\frac{15(x+2)}{6(x-2)(x+2)}=\frac{15(x+2)}{6(x^2-4)}\)

\(\frac{1}{6-3x}=\frac{x+2}{3(2-x)}=\frac{2(x+2)^2}{6(2-x)(2+x)}=\frac{-2(x+2)^2}{6(x^2-4)}\)

b.

\(\frac{1}{x+2}=\frac{x(2-x)}{x(x+2)(2-x)}=\frac{x(2-x)}{x(4-x^2)}\)

\(\frac{8}{2x-x^2}=\frac{8(x+2)}{(x+2)x(2-x)}=\frac{8(x+2)}{x(4-x^2)}\)

c.

\(\frac{4x^2-3x+5}{x^3-1}\)

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

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

a) Ta có: \(\left|2x-\dfrac{1}{3}\right|\ge0\forall x\)

\(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|-\dfrac{7}{4}\ge-\dfrac{7}{4}\forall x\)

Dấu '=' xảy ra khi \(2x=\dfrac{1}{3}\)

hay \(x=\dfrac{1}{6}\)

Vậy: \(A_{min}=-\dfrac{7}{4}\) khi \(x=\dfrac{1}{6}\)

b) Ta có: \(\dfrac{1}{3}\left|x-2\right|\ge0\forall x\)

\(\left|3-\dfrac{1}{2}y\right|\ge0\forall y\)

Do đó: \(\dfrac{1}{3}\left|x-2\right|+\left|3-\dfrac{1}{2}y\right|\ge0\forall x,y\)

\(\Leftrightarrow\dfrac{1}{3}\left|x-2\right|+\left|3-\dfrac{1}{2}y\right|+4\ge4\forall x,y\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-2=0\\3-\dfrac{1}{2}y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=6\end{matrix}\right.\)

Vậy: \(B_{min}=4\) khi x=2 và y=6

Câu 1:

\(\left(x-2\right)\left(x^2+2x+4\right)+25x=x\left(x+5\right)\left(x-5\right)+8\)

\(\Leftrightarrow x^3-8+25x=x\left(x^2-25\right)+8\)

\(\Leftrightarrow x^3-8+25x=x^3-25x+8\)

\(\Leftrightarrow x^3-8+25x-x^3+25x-8=0\)

\(\Leftrightarrow50x-16=0\)

\(\Leftrightarrow50x=16\)

\(\Leftrightarrow x=\dfrac{8}{25}\)

Câu 2 :

\(\dfrac{x+5}{4}+\dfrac{3+2x}{3}=\dfrac{6x-1}{3}-\dfrac{1-2x}{12}\)

<=> \(\dfrac{3\left(x+5\right)}{12}+\dfrac{4\left(3+2x\right)}{12}=\dfrac{4\left(6x-1\right)}{12}-\dfrac{1-2x}{12}\)

<=>\(\dfrac{3x+15+12+8x}{12}=\dfrac{24x-4-1+2x}{12}\)

<=> 3x + 15 + 12 + 8x = 24x - 4 - 1 +2x

<=> 11x+27 = 26x -5

<=> ( 26x - 5 ) - ( 11x + 27 ) = 0

<=> 15x - 32 = 0

<=> 15x = 32

<=> x = \(\dfrac{32}{15}\)

Câu 3:

x - 4/3 - 3x - 1/12 = 3x + 1/4 + 9x - 2/8

<=> 4x - 16 - 3x + 1/12 = 6x + 2 + 9x - 2/8

<=> x - 15/12 = 15x/8

<=> 8x - 120 = 180x

<=> 120 = -172x <=> x = -172/120 = -43/30

Mình nghĩ ra câu C rồi bạn nào giúp mình nghĩ nốt câu A,B hộ mình nhé mình cảm ơn!

a:6x-5-9x^2

=-(9x^2-6x+5)

=-(9x^2-6x+1+4)

=-(3x-1)^2-4<=-4

=>A>=2/-4=-1/2

Dấu = xảy ra khi x=1/3

b: \(B=\dfrac{4x^2-6x+4-1}{2x^2-3x+2}=2-\dfrac{1}{2x^2-3x+2}\)

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

=2(x^2-2*x*3/4+9/16+7/16)

=2(x-3/4)^2+7/8>=7/8

=>-1/2x^2-3x+2<=-1:7/8=-8/7

=>B<=-8/7+2=6/7

Dâu = xảy ra khi x=3/4

đk: x khác -1; 3

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

<=> \(\dfrac{x}{2\left(x+1\right)}-\dfrac{2x}{\left(x-3\right)\left(x+1\right)}+\dfrac{x}{2\left(x-3\right)}=0\)

<=> \(\dfrac{x\left(x-3\right)-4x+x\left(x+1\right)}{2\left(x+1\right)\left(x-3\right)}=0\)

<=> \(x^2-3x-4x+x^2+x=0\)

<=> 2x² - 6x = 0

<=> 2x(x-3) = 0

Mà x khác 3

<=> 2x = 0

<=> x = 0

\(\dfrac{2\text{x}-1}{3}=\dfrac{3\text{x}+1}{4}\)

\(\Leftrightarrow=\dfrac{4\left(2\text{x}-1\right)}{12}=\dfrac{3\left(3\text{x}+1\right)}{12}\)

\(\Leftrightarrow8\text{x}-4=9\text{x}+3\)

\(\Leftrightarrow8\text{x}-9\text{x}=3+4\)

\(\Leftrightarrow-x=7\)

\(\Leftrightarrow x=-7\)

`(2x-1)/3 = (3x+1)/4`

`=> (2x-1).4= 3.(3x+1)`

`=> 8x -4= 9x+3`

`=> 8x-9x =3+4`

`=> -x=7`

`=>x=-7`