\(a,ĐK:...\\ PT\Leftrightarrow x^2-6x=x^2-7x+10\\ \Leftrightarrow x=10\left(tm\right)\\ b,ĐK:...\\ PT\Leftrightarrow2x\left(4-x\right)-\left(2-2x\right)\left(8-x\right)=\left(8-x\right)\left(4-x\right)\\ \Leftrightarrow8x-2x^2+16+18x-2x^2=32-12x+x^2\\ \Leftrightarrow3x^2-38x+16=0\left(casio\right)\\ c,ĐK:...\\ PT\Leftrightarrow2x\left(x-4\right)-4x=0\\ \Leftrightarrow2x^2-12x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)

GHI RÕ DÙM MÌNH ĐK CỦA CẢ 3 CÂU LUÔN ĐC KO Á.

1: \(\Leftrightarrow x^2-6x=x^2-7x+10\)

hay x=10

1: \(\Leftrightarrow x^2-6x=x^2-7x+10\)

hay x=10

sao câu 1 hoài v ạ.Còn câu 2,3 nữa á.

a: ĐKXĐ: x∈R

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

=>\(\frac{5\left(x^2-2x+5\right)-8\left(x^2-2x+2\right)}{\left(x^2-2x+2\right)\left(x^2-2x+5\right)}=3\)

=>\(3\left(x^2-2x+2\right)\left(x^2-2x+5\right)=5x^2-10x+25-8x^2+16x-16=-3x^2+6x+9\)

=>\(3\left\lbrack\left(x^2-2x\right)^2+7\left(x^2-2x\right)+10\right\rbrack=-3\left(x^2-2x\right)+9\)

=>\(\left(x^2-2x\right)^2+7\left(x^2-2x\right)+10=-\left(x^2-2x\right)+3\)

=>\(\left(x^2-2x\right)^2+8\left(x^2-2x\right)+7=0\)

=>\(\left(x^2-2x+1\right)\left(x^2-2x+7\right)=0\)

=>\(x^2-2x+1=0\)

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

=>x-1=0

=>x=1(nhận)

b: ĐKXĐ: x<>0

\(\frac{x^2-4x+3}{2x}+\frac{x^2+12x+3}{x^2+3}=4\)

=>\(\frac{x^2+3}{2x}-2+\frac{12x}{x^2+3}+1=4\)

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

=>\(\frac{\left(x^2+3\right)^2+24x^2}{2x\left(x^2+3\right)}=5\)

=>\(\left(x^2+3\right)^2+24x^2-10x\left(x^2+3\right)=0\)

=>\(\left(x^2+3\right)^2-4x\left(x^2+3\right)-6x\left(x^2+3\right)+24x^2=0\)

=>\(\left(x^2+3\right)\left(x^2+3-4x\right)-6x\left(x^2+3-4x\right)=0\)

=>\(\left(x^2-6x+3\right)\left(x^2-4x+3\right)=0\)

TH1: \(x^2-6x+3=0\)

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

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

=>\(\left[\begin{array}{l}x-3=\sqrt6\\ x-3=-\sqrt6\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\sqrt6+3\left(nhận\right)\\ x=-\sqrt6+3\left(nhận\right)\end{array}\right.\)

TH2: \(x^2-4x+3=0\)

=>\(x^2-x-3x+3=0\)

=>(x-1)(x-3)=0

=>x=1(nhận) hoặc x=3(nhận)

điều kiện xác định \(x\ne0\)

ta có : \(\dfrac{x+1}{x^2+2x+4}-\dfrac{x-2}{x^2-2x+4}=\dfrac{6}{x\left(x^4+4x^2+16\right)}\)

\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x^2-2x+4\right)-\left(x-2\right)\left(x^2+2x+4\right)}{\left(x^2+2x+4\right)\left(x^2-2x+4\right)}=\dfrac{6}{x\left(x^4+4x^2+16\right)}\)

\(\Leftrightarrow-x^2+2x+12=\dfrac{6}{x}\Leftrightarrow x\left(-x^2+2x+12\right)=6\)

\(\Leftrightarrow-x^3+2x^2+12x=6\Leftrightarrow-x^3+2x^2+12x-6=0\)

tới đây bn bấm máy tính nha

câu b lm tương tự nha

a) Sửa đề: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)

ĐKXĐ: \(x\notin\left\{3;\dfrac{1}{5}\right\}\)

Ta có: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)

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

Suy ra: \(9-3x+10x-2=4\)

\(\Leftrightarrow7x+7=4\)

\(\Leftrightarrow7x=-3\)

hay \(x=-\dfrac{3}{7}\)

Vậy: \(S=\left\{-\dfrac{3}{7}\right\}\)

b: \(\Leftrightarrow\dfrac{20}{x}-\dfrac{20}{x+20}=\dfrac{1}{6}\)

=>\(\dfrac{20x+400-20x}{x\left(x+20\right)}=\dfrac{1}{6}\)

=>x*(x+20)=400*6=2400

=>x^2+20x-2400=0

=>(x+60)(x-40)=0

=>x=-60 hoặc x=40

c: \(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}=\dfrac{8}{4x^2-1}\)

=>(2x+1)^2-(2x-1)^2=8

=>4x^2+4x+1-4x^2+4x-1=8

=>8x=8

=>x=1(nhận)

câu b sai đề rồi anh ơi và câu a đâu rồi ạ

a: ĐKXĐ: x<>-2/3

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

=>5(3x+2)=2x+1

=>15x+10=2x+1

=>13x=-9

=>\(x=-\frac{9}{13}\) (nhận)

b: ĐKXĐ: x∉{1;3}

\(\frac{2x^2-5x+2}{x-1}=\frac{2x^2+x+15}{x-3}\)

=>\(\left(2x^2-5x+2\right)\left(x-3\right)=\left(2x^2+x+15\right)\left(x-1\right)\)

=>\(2x^3-6x^2-5x^2+15x+2x-6=2x^3-2x^2+x^2-x+15x-15\)

=>\(-11x^2+17x-6=-x^2+14x-15\)

=>\(-10x^2+3x+9=0\)

=>\(10x^2-3x-9=0\)

=>\(x^2-\frac{3}{10}x-\frac{9}{10}=0\)

=>\(x^2-2\cdot x\cdot\frac{3}{20}+\frac{9}{400}-\frac{9}{400}-\frac{9}{10}=0\)

=>\(\left(x-\frac{3}{20}\right)^2=\frac{9}{400}+\frac{9}{10}=\frac{9}{400}+\frac{360}{400}=\frac{369}{400}\)

=>\(x-\frac{3}{20}=\pm\frac{3\sqrt{41}}{20}\)

=>\(\left[\begin{array}{l}x=\frac{3\sqrt{41}+3}{20}\left(nhận\right)\\ x=\frac{-3\sqrt{41}+3}{20}\left(nhận\right)\end{array}\right.\)

c: ĐKXĐ: x∉{3;-3}

\(\frac{2x+3}{x-3}-\frac{4}{x+3}=\frac{24}{x^2-9}+2\)

=>\(\frac{\left(2x+3\right)\left(x+3\right)-4\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{24+2\left(x^2-9\right)}{\left(x-3\right)\left(x+3\right)}\)

=>(2x+3)(x+3)-4(x-3)=\(24+2x^2-18\)

=>\(2x^2+6x+3x+9-4x+12=2x^2+6\)

=>5x+21=6

=>5x=-15

=>x=-3(loại)

1: Sửa đề: 2/x+2

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

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

=>4x-3=-3x-6

=>7x=-3

=>x=-3/7(nhận)

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

=>9x-3x^2+3-x+3-9x+x-3x^2=2(3x-1)(x-3)

=>-6x^2+6=2(3x^2-10x+3)

=>-6x^2+6=6x^2-20x+6

=>-12x^2+20x=0

=>-4x(3x-5)=0

=>x=5/3(nhận) hoặc x=0(nhận)

3: \(\Leftrightarrow x\cdot\dfrac{8}{3}-\dfrac{2}{3}=1+\dfrac{5}{4}-\dfrac{1}{2}x\)

=>x*19/6=35/12

=>x=35/38