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

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

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

\(\Leftrightarrow2x^2-2x-5x+5-3x^2+6x+5x-10=-x^2+x+2x-2\)

\(\Leftrightarrow2x^2-2x-5x+5-3x^2+6x+5x-10+x^2-x-2x+2=0\)

\(\Leftrightarrow x=0\)

Vậy tập nghiệm của pt là S={0}

E muốn show cách mới nghĩ ra nhưng sợ sai nên e lm cách này cho chắc 

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

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

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

\(\Leftrightarrow-x^2+4x-5=-1\)

\(\Leftrightarrow x^2-4x+4=0\)

\(\Leftrightarrow\left(x-2\right)^2=0\)

\(\Leftrightarrow x-2=0\)

\(\Leftrightarrow x=2\)

Vậy pt cs nghiệm { 2 }

BD=√AB2+AD2=√62+82=10  (cm)

BE=√AB2+AE2=√62+62=6√2 (cm)

ED=√AD2+AE2=√82+62=10)

⇒⇒ Chu vi ΔBED=BD+BE+ED=20+6√2≈28,49 (cm)

\(\frac{2x}{x-2}-\frac{5}{x+2}=\frac{x^2+12}{x^2-4}\)

\(\Leftrightarrow\frac{2x\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{x^2+12}{\left(x+2\right)\left(x-2\right)}\)

\(\Rightarrow2x\left(x+2\right)-5\left(x-2\right)=x^2+12\)

\(\Leftrightarrow2x^2+4x-5x+10=x^2+12\)

\(\Leftrightarrow2x^2+4x-5x+10-x^2-12=0\)

\(\Leftrightarrow x^2-x-2=0\)

\(\Leftrightarrow x^2+x-2x-2=0\)

\(\Leftrightarrow\left(x^2+x\right)-\left(2x+2\right)=0\)

\(\Leftrightarrow x\left(x+1\right)-2\left(x+1\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)

\(\orbr{\begin{cases}x-2=0\Leftrightarrow x=2\\x+1=0\Leftrightarrow x=-1\end{cases}}\)

Vậy tập nghiệm của phương trình là S={2;-1}

ĐKXĐ: \(x\ne\pm2\)

\(\frac{2x}{x-2}-\frac{5}{x+2}=\frac{x^2+12}{x^2-4}\)

<=> \(\frac{2x}{x-2}-\frac{5}{x+2}=\frac{x^2+12}{\left(x-2\right)\left(x+2\right)}\)

<=> 2x(x + 2) - 5(x - 2) = x² + 12

<=> 2x² + 4x - 5x + 10 = x² + 12

<=> 2x² - x + 10 = x² + 12

<=> 2x² - x + 10 - x² - 12 = 0

<=> x² - x - 2 = 0

<=> (x - 2)(x + 1) = 0

<=> x - 2 = 0 hoặc x + 1 = 0

<=> x = 2 (ktm) hoặc x = -1 (tm)

=> x = -1

       x(x-1)=x(x+3)

<=> x^2-x=x^2+3x

<=> x^2-x-x^2-3x=0

<=> -4x=0

<=> x=0

       (x-1)(x+3)=x^2-4

<=> x^2+3x-x-3=x^2-4

<=> x^2+2x-3=x^2-4 

<=> x^2-x^2+2x=-4+3

<=> 2x=-1

<=> x=-1/2

       (x-2)(x-5)=(x-3)(x-4)

<=> x^2-7x+10=x^2-7x+12

<=> x^2-7x-x^2+7x=12-10

<=> 0x=2(vô nghiệm)

1.Đ

2.S

3.S

4.Đ

5.S

6.Đ

a) ĐKXĐ: \(x\ne0;x\ne5\)

b) \(P=\frac{x^2+2x}{2x+10}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}\)

\(P=\frac{x^2+2x}{2\left(x+5\right)}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}\)

\(P=\frac{x\left(x^2+2x\right)}{2x\left(x+5\right)}+\frac{2\left(x+5\right)\left(x-5\right)}{2x\left(x+5\right)}+\frac{50-5x}{2x\left(x+5\right)}\)

\(P=\frac{x\left(x^2+2x\right)+2\left(x+5\right)\left(x-5\right)+50-5x}{2x\left(x+5\right)}\)

\(P=\frac{x\left(x^2-5+4x\right)}{2x\left(x+5\right)}\)

\(P=\frac{x^2-5x+4}{2\left(x+5\right)}\)

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

\(P=\frac{x-1}{2}\)

c) +) P = 0

\(\frac{x-1}{2}=0\)

<=> x - 1 = 0

<=> x = 1

+) P = 1/4 

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

<=> 4(x - 1) = 2.1

<=> 4x - 4 = 2

<=> 4x = 2 + 4

<=> 4x = 6

<=> x = 6/4 = 3/2

x²+2x-4=-12+3x+x²

<=> x²-x²+2x-3x=-12+4 (chuyển vế đổi dấu)

<=>-x=-8

<=>x=8

nhớ k cho mik đó nhoa

Đ/án 5/14 nha bạn !!!!!!!!!!!!!!!!!