TK

https://lazi.vn/edu/exercise/giai-phuong-trinh-4x-5-x-1-2-x-x-1-7-x-2-3-x-5

a: \(\Leftrightarrow4x-5=2x-2+x\)

=>4x-5=3x-2

=>x=3(nhận)

b: =>7x-35=3x+6

=>4x=41

hay x=41/4(nhận)

c: \(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{x+2}{x-4}=\dfrac{-3}{2\left(x-4\right)}-\dfrac{5}{6}\)

\(\Leftrightarrow\dfrac{28}{6\left(x-4\right)}-\dfrac{6\left(x+2\right)}{6\left(x-4\right)}=\dfrac{-9}{6\left(x-4\right)}-\dfrac{5\left(x-4\right)}{6\left(x-4\right)}\)

\(\Leftrightarrow28-6x-12=-9-5x+20\)

=>-6x+16=-5x+11

=>-x=-5

hay x=5(nhận)

d: \(\Leftrightarrow x^2+2x+1-\left(x^2-2x+1\right)=16\)

\(\Leftrightarrow4x=16\)

hay x=4(nhận)

a: Ta có: \(4x-2\left(1-x\right)=5\left(x-4\right)\)

\(\Leftrightarrow4x-2+2x=5x-20\)

\(\Leftrightarrow x=-18\)

b: Ta có: \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)

\(\Leftrightarrow6x+4\left(1-3x\right)=3\left(-x+1\right)\)

\(\Leftrightarrow6x+4-12x=-3x+3\)

\(\Leftrightarrow-3x=-1\)

hay \(x=\dfrac{1}{3}\)

c: Ta có: \(\left(x+2\right)^2-3\left(x+2\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)

d: Ta có: \(\dfrac{x-5}{x}+\dfrac{x-3}{x+5}=\dfrac{x-25}{x\left(x+5\right)}\)

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

\(\Leftrightarrow2x^2-4x=0\)

\(\Leftrightarrow2x\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=2\left(nhận\right)\end{matrix}\right.\)

\(\dfrac{1}{x^2-3x+2}+\dfrac{1}{x^2-5x+6}-\dfrac{2}{x^2-4x+3}\) = 0  (ĐKXĐ: x \(\ne\) 1; x \(\ne\) 2; x \(\ne\) 3)

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

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

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

Vậy pt trên có vô số nghiệm và x \(\ne\) 1; x \(\ne\) 2; x \(\ne\) 3

Chúc bn học tốt!

3x+5

=0

\(pt\text{ tương đương:}\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}-\dfrac{2}{\left(x-1\right)\left(x-3\right)}=0\text{ tương đương với: }\dfrac{1}{x-1}-\dfrac{1}{x-2}+\dfrac{1}{x-2}-\dfrac{1}{x-3}-\dfrac{2}{\left(x-1\right)\left(x-3\right)}=0\text{ hay:}\dfrac{1}{x-1}-\dfrac{1}{x-3}-\dfrac{1}{x-1}+\dfrac{1}{x-3}=0\text{ đúng}\)

(1x+3−12x+2

\(\dfrac{4x}{x^2+4x+3}-1=6\left(\dfrac{1}{x+3}-\dfrac{1}{2x+2}\right)\left(x\ne-3;x\ne-1\right)\\ < =>\dfrac{4x}{x^2+4x+4-1}-1=\dfrac{6}{x+3}-\dfrac{6}{2x+2}\\ < =>\dfrac{4x}{\left(x+2\right)^2-1}-1=\dfrac{6}{x+3}-\dfrac{6}{2\left(x+1\right)}\\ < =>\dfrac{4x}{\left(x+1\right)\left(x+3\right)}-1=\dfrac{6}{x+3}-\dfrac{3}{x+1}\)

suy ra:

`4x-(x+1)(x+3)=6(x+1)-3(x+3)`

\(< =>4x-\left(x^2+3x+x+3\right)=6x+6-3x-9\)

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

\(< =>-x^2+4x-4x-6x+3x-3-6+9=0\)

\(< =>-x^2-3x=0\\ < =>x^2+3x=0\\ < =>x\left(x+3\right)=0\\ < =>\left[{}\begin{matrix}x=0\\x+3=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\left(tm\right)\\x=-3\left(ktmđk\right)\end{matrix}\right.\)

vậy pt có tập nghiệm S={0}

PT 2 

\(\Leftrightarrow\dfrac{3}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}+\dfrac{2x}{\left(x-2\right)\left(x-3\right)}-\dfrac{1}{\left(x-1\right)\left(x-2\right)}=0\) ( \(x\ne1;x\ne2;x\ne3\))

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

\(\Rightarrow2x^2-3x+6=0\)

=> PT vô nghiệm.

\(25\sqrt{\dfrac{x-3}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\left(x\ge3\right)\)

\(=25\sqrt{\dfrac{1}{25}.\left(x-3\right)}-7\sqrt{\dfrac{4}{9}.\left(x-3\right)}-7\sqrt{x^2-9}+18\sqrt{\dfrac{1}{9}.\left(x^2-9\right)}=0\)

\(=5\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)

\(\Rightarrow\dfrac{1}{3}\sqrt{x-3}-\sqrt{\left(x-3\right)\left(x+3\right)}=0\Rightarrow\sqrt{x-3}-3\sqrt{\left(x-3\right)\left(x+3\right)}=0\)

\(\Rightarrow\sqrt{x-3}\left(1-3\sqrt{x+3}\right)=0\Rightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\1=3\sqrt{x+3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{26}{9}\left(l\right)\end{matrix}\right.\)

a: Ta có: \(6-4x=5(x+3)+3\)

\(\Leftrightarrow6-4x-5x-12-3=0\)

\(\Leftrightarrow-9x=9\)

hay x=-1

b: Ta có: \(\dfrac{x+3}{2}-1=\dfrac{x-1}{3}+\dfrac{x+5}{6}\)

\(\Leftrightarrow15x+45-30=10x-30+5x+25\)

\(\Leftrightarrow15=-5\left(loại\right)\)

c: Ta có: \(\left(x-2\right)\left(2x+1\right)-3\left(x-2\right)=0\)

\(\Leftrightarrow2\left(x-2\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)

d: Ta có: \(\dfrac{2}{x^2-2x}+\dfrac{1}{x}=\dfrac{x+2}{x-2}\)

\(\Leftrightarrow2+x-2=x^2+2x\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)

b: Ta có: \(\sqrt{x^2-6x+9}-\dfrac{\sqrt{6}+\sqrt{3}}{\sqrt{2}+1}=0\)

\(\Leftrightarrow x^2-6x+9=3\)

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

\(\text{Δ}=\left(-6\right)^2-4\cdot1\cdot6=36-24=12\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{6-2\sqrt{3}}{2}=3-\sqrt{3}\\x_2=3+\sqrt{3}\end{matrix}\right.\)

ĐKXD phức tạp nên ko tìm ngay

Đặt \(x^2=t>0\Rightarrow\sqrt{12-\dfrac{3}{t}}+\sqrt{4t-\dfrac{3}{t}}=4t\)

Đặt \(\sqrt{4t-\dfrac{3}{t}}=a\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{t}=4t-a^2\\3=4t^2-ta^2\end{matrix}\right.\)

\(\Rightarrow\sqrt{4\left(4t^2-ta^2\right)-\left(4t-a^2\right)}+a=4t\)

\(\Rightarrow\sqrt{16t^2-4ta^2-4t+a^2}=4t-a\)

\(\Rightarrow16t^2-4ta^2-4t+a^2=\left(4t-a\right)^2\)

\(\Rightarrow16t^2-4ta^2-4t+a^2=16t^2-8ta+a^2\)

\(\Rightarrow4ta^2-8ta+4t=0\)

\(\Rightarrow4t\left(a-1\right)^2=0\Rightarrow\left[{}\begin{matrix}t=0\left(loại\right)\\a=1\end{matrix}\right.\)

\(\Rightarrow\sqrt{4t-\dfrac{3}{t}}=1\Rightarrow4t^2-t-3=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-\dfrac{3}{4}< 0\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow x^2=1\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

Thử lại 2 nghiệm vào pt ban đầu đều thỏa mãn