b)

ĐKXĐ: \(x\notin\left\{2;3;\dfrac{1}{2}\right\}\)

Ta có: \(\dfrac{x+4}{2x^2-5x+2}+\dfrac{x+1}{2x^2-7x+3}=\dfrac{2x+5}{2x^2-7x+3}\)

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

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

Suy ra: \(x^2-3x+4x-12+x^2-2x+x-2=2x^2-4x+5x-10\)

\(\Leftrightarrow2x^2-14=2x^2+x-10\)

\(\Leftrightarrow2x^2-14-2x^2-x+10=0\)

\(\Leftrightarrow-x-4=0\)

\(\Leftrightarrow-x=4\)

hay x=-4(nhận)

Vậy: S={-4}

a: \(2^{x^2-2x+1}=1\)

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

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

=>x-1=0

=>x=1

b: \(7^{x^2+7x}=5764801\)

=>\(7^{x^2+7x}=7^8\)

=>\(x^2+7x=8\)

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

=>(x+8)(x-1)=0

=>\(\left[{}\begin{matrix}x+8=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=1\end{matrix}\right.\)

c: \(6^{x^2+12x}=6^{7x}\)

=>\(x^2+12x=7x\)

=>\(x^2+5x=0\)

=>x(x+5)=0

=>\(\left[{}\begin{matrix}x=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

d: \(\left(\dfrac{1}{3}\right)^{x-1}=3^{2x-5}\)

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

=>-x+1=2x-5

=>-x-2x=-5-1

=>-3x=-6

=>x=2

e: \(\left(\dfrac{1}{5}\right)^{3x+5}=5^{2x+1}\)

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

=>-3x-5=2x+1

=>-5x=6

=>\(x=-\dfrac{6}{5}\)

1.

ĐK: \(x\ne7;x\ne-1;x\ne3\)

\(\dfrac{2x-5}{x^2-6x-7}\le\dfrac{1}{x-3}\left(1\right)\)

TH1: \(x< -1\)

\(\left(1\right)\Leftrightarrow\left(2x-5\right)\left(x-3\right)\ge x^2-6x-7\)

\(\Leftrightarrow2x^2-11x+15\ge x^2-6x-7\)

\(\Leftrightarrow x^2-5x+22\ge0\)

\(\Leftrightarrow\) Bất phương trình đúng với mọi \(x< -1\)

TH2: \(-1< x< 3\)

\(\left(1\right)\Leftrightarrow\left(3-x\right)\left(2x-5\right)\ge\left(7-x\right)\left(x+1\right)\)

\(\Leftrightarrow-2x^2+11x-15\ge-x^2+6x+7\)

\(\Leftrightarrow-x^2+5x-22\ge0\)

\(\Rightarrow\) vô nghiệm

TH3: \(3< x< 7\)

Khi đó \(\dfrac{2x-5}{x^2-6x-7}\le0\); \(\dfrac{1}{x-3}>0\)

\(\Rightarrow\) Bất phương trình đúng với mọi \(3< x< 7\)

TH4: \(x>7\)

\(\left(1\right)\Leftrightarrow\left(2x-5\right)\left(x-3\right)\le x^2-6x-7\)

\(\Leftrightarrow2x^2-11x+15\le x^2-6x-7\)

\(\Leftrightarrow x^2-5x+22\le0\)

\(\Rightarrow\) vô nghiệm

Vậy ...

Các bài kia tương tự, chứ giải ra mệt lắm.

a: \(\Leftrightarrow7\left(7-3x\right)+12\left(5x+2\right)=84\left(x+13\right)\)

\(\Leftrightarrow49-21x+60x+24=84x+1092\)

\(\Leftrightarrow39x-84x=1092-73\)

=>-45x=1019

hay x=-1019/45

b: \(\Leftrightarrow21\left(x+3\right)-14=4\left(5x+9\right)-7\left(7x-9\right)\)

=>21x+63-14=20x+36-49x+63

=>21x+49=-29x+99

=>50x=50

hay x=1

c: \(\Leftrightarrow7\left(2x+1\right)-3\left(5x+2\right)=21x+63\)

=>14x+7-15x-6-21x-63=0

=>-22x-64=0

hay x=-32/11

d: \(\Leftrightarrow35\left(2x-3\right)-15\left(2x+3\right)=21\left(4x+3\right)-17\cdot105\)

=>70x-105-30x-45=84x+63-1785

=>40x-150-84x+1722=0

=>-44x+1572=0

hay x=393/11

a, msc 12.7=84 

Chuyển vế về =0 rồi làm

b,msc 28

c,làm tương tự

a, \(\Rightarrow49-21x+60x+24=84x+1092\)

\(\Leftrightarrow-45x=1019\Leftrightarrow x=-\dfrac{1019}{45}\)

b, \(\Rightarrow21\left(x+3\right)-14=4\left(5x+9\right)-7\left(7x-9\right)\)

\(\Leftrightarrow21x+63-14=20x+36-49x+63\)

\(\Leftrightarrow50x=50\Leftrightarrow x=1\)

c, \(\Rightarrow14x+7-15x-6=21x+63\Leftrightarrow-22x=62\Leftrightarrow x=-\dfrac{31}{11}\)

d, \(\Rightarrow35\left(2x-3\right)-15\left(2x+3\right)=21\left(4x+3\right)-105.17\)

\(\Leftrightarrow70x-105-30x-45=84x+63-1785\)

\(\Leftrightarrow-44x=-1572\Leftrightarrow x=\dfrac{393}{11}\)

\(\Leftrightarrow\dfrac{x+5}{2x^2-5x+2}=\dfrac{2x+7-x-2}{2x^2-7x+3}\)

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

\(\Leftrightarrow\left(x+5\right)\left(x-3-x+2\right)=0\)

=>x+5=0

hay x=-5

\(i.\dfrac{\left(2x+1\right)^2}{5}-\dfrac{\left(x-1\right)^2}{3}=\dfrac{7x^2-14x-5}{15}\)

\(\Leftrightarrow\dfrac{4x^2+4x+1}{5}-\dfrac{x^2-2x+1}{3}=\dfrac{7x^2-14x-5}{15}\)

\(\Leftrightarrow\dfrac{12x^2+12x+3}{15}-\dfrac{5x^2-10x+5}{15}=\dfrac{7x^2-14x-5}{15}\)

\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5=7x^2-14x-5\)

\(\Leftrightarrow36x=-3\)

\(\Leftrightarrow x=-\dfrac{1}{12}\)

\(k.x+\dfrac{2x+\dfrac{x-1}{5}}{3}=1-\dfrac{3x-\dfrac{1-2x}{3}}{5}\)

\(\Leftrightarrow\dfrac{15x}{15}+\dfrac{10x+x-1}{15}=\dfrac{15}{15}-\dfrac{9x-1+2x}{15}\)

\(\Leftrightarrow15x+9x-1=14-7x\)

\(\Leftrightarrow31x=15\)

\(\Leftrightarrow x=\dfrac{15}{31}\)

câu h tương tự bài trước câu g mik làm nhé

Lời giải:

ĐKXĐ:.....

Ta có: \(\frac{x+5}{x(2x-5)+2}+\frac{x+2}{x(2x-7)+3}=\frac{2x+7}{2x^2-7x+3}\)

\(\Leftrightarrow \frac{x+5}{x(2x-5)+2}=\frac{2x+7}{2x^2-7x+3}-\frac{x+2}{2x^2-7x+3}\)

\(\Leftrightarrow \frac{x+5}{x(2x-5)+2}=\frac{2x+7-(x+2)}{2x^2-7x+3}\)

\(\Leftrightarrow \frac{x+5}{x(2x-5)+2}=\frac{x+5}{2x^2-7x+3}\)

\(\Leftrightarrow (x+5)\left(\frac{1}{x(2x-5)+2}-\frac{1}{2x^2-7x+3}\right)=0\)

TH1: \(x+5=0\Leftrightarrow x=-5\) (thỏa mãn)

TH2: \(\frac{1}{x(2x-5)+2}-\frac{1}{2x^2-7x+3}=0\)

\(\Leftrightarrow x(2x-5)+2=2x^2-7x+3\)

\(\Leftrightarrow -5x+2=-7x+3\)

\(\Leftrightarrow 2x=1\Leftrightarrow x=\frac{1}{2}\). Thử lại thây mẫu số bằng 0 (không thỏa mãn ĐKXĐ)
Vậy PT có nghiệm duy nhất \(x=-5\)

ĐỀ BÀI LÀ GÌ VẬY BẠN.