\(\frac{x-3}{5}-\frac{2x-1}{10}=\frac{x+1}{2}+\frac{1}{4}\)

\(< =>\frac{\left(x-3\right).4}{20}-\frac{\left(2x-1\right).2}{20}=\frac{\left(x+1\right).10}{20}+\frac{5}{20}\)

\(< =>4x-12-4x+2=10x+10+5\)

\(< =>10x=-10-10-5=-25\)

\(< =>x=-\frac{25}{10}=-\frac{5}{2}\)

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

\(< =>\frac{\left(x+3\right).15}{30}-\frac{\left(2x-1\right).10}{30}-\frac{30}{30}=\frac{\left(x+5\right).5}{30}\)\(< =>15x+45-20x+10-30=5x+25\)

\(< =>-5x+25=5x+25< =>10x=0< =>x=0\)

\(\frac{x-1}{4}-\frac{5-2x}{9}=3x-\frac{2}{3}\)

\(< =>\frac{\left(x-1\right).9}{36}-\frac{\left(5-2x\right).4}{36}=\frac{3x.36}{36}-\frac{2.12}{36}\)

\(< =>\left(x-1\right).9-\left(5-2x\right).4=108x-24\)

\(< =>9x-9-20+8x=108x-24\)

\(< =>108x-17x=-29+24\)

\(< =>91x=-5< =>x=-\frac{5}{91}\)

Đây là lớp 8 nha các b giúp mk với

Do mk viết nhầm

a. 2x

b.\({3x}\over x^2-1\)

a, 2(4x - 7 ) = 3(x + 1) + 18

⇌ 8x -14 = 3x + 3 + 18

⇌ 5x = 35 ⇌ x = 7

→ S = \(\left\{7\right\}\)

b, ( 2x - 1 )² - 4x ( x - 3 ) = -11

⇌ 4x² - 2x + 1 - 4x² + 12 = -11

⇌ 10x = -12

⇌ x = \(-\frac{12}{10}\)

→ S = \(\left\{-\frac{12}{10}\right\}\)

c, ( 2x - 5 )² - ( x + 2 )² = 0

⇌ ( 2x - 5 -x + 2 )² = 0

⇌ ( x - 3 )² = 0

⇌ x - 3 = 0 ⇌ x = 3

→ S = \(\left\{3\right\}\)

d, ( x - 6 ) ( x + 1 ) = 2(x + 1)

⇌ ( x - 6 - 2 ) ( x+ 1) = 0

⇌ x² - 7x - 8 =0

⇌ ( x - 8 ) ( x + 1 ) = 0

⇒\(\left\{{}\begin{matrix}x-8=0\\x+1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=8\\x=-1\end{matrix}\right.\)

→ S = \(\left\{8;-1\right\}\)

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

⇌ 5( x - 3) = 20 - 2(1 - 2x)

⇌ 5x - 4x = 15 + 20 + 2

⇌ x = 37

→ S = \(\left\{37\right\}\)

g, \(\frac{3x+2}{2}+\frac{5-2x}{3}=\frac{11}{6}\)

⇌ 3(3x + 2) + 2(5 - 2x) = 11

⇌ 6x + 6 + 10 - 4x = 11

⇌ 2x = -5

⇌ x = \(-\frac{5}{2}\)

→ S = \(\left\{-\frac{5}{2}\right\}\)

h, \(\frac{x-2}{x+2}-\frac{3}{x-2}=\frac{9x-66}{x^2-4}\)

⇌ (x - 2)² - 3(x - 2) = 9x - 66

⇌ x² - 4x + 4 - 3x - 6 = 9x - 66

⇌ x² -16 + 64 = 0

⇌ (x - 8)² = 0

⇌ x - 8 = 0

⇌ x = 8

→ S = \(\left\{8\right\}\)

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

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

\(\Rightarrow\)\(x^2+3x+2+x^2-2x-3=2+x^2+4x+3\)

\(\Rightarrow x^2-3x-6=0\)

.....

\(\frac{x+1}{x-2}+\frac{2x-1}{x-1}=\frac{2}{x^2-3x+2}+\frac{11}{2}\)

\(\Rightarrow\frac{2\left(x+1\right)\left(x-1\right)}{2\left(x-2\right)\left(x-1\right)}+\frac{2\left(2x-1\right)\left(x-2\right)}{2\left(x-1\right)\left(x-2\right)}\)\(=\frac{4}{2\left(x-1\right)\left(x-2\right)}+\frac{22\left(x-1\right)\left(x-2\right)}{2\left(x-1\right)\left(x-2\right)}\)

\(\Rightarrow2x^2-2+4x^2-10x+4=4+22x^2-66x+44\)

.....

\(3,\)\(-2x^2+4x+3\)

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

\(=-2\left[\left(x^2-2x+1\right)-\frac{5}{2}\right]\)

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

Đa thức này lớn nhất =5 khi và chỉ khi \(\left(x-1\right)^2\)nhỏ nhất 

\(\Rightarrow x-1=0\)

\(\Rightarrow x=1\)

Đây là những bài cơ bản mà bạn!

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

\(< =>\frac{\left(5x-2\right).2}{6}=\frac{\left(5-3x\right).3}{6}\)

\(< =>\left(5x-2\right).2=\left(5-3x\right).3\)

\(< =>10x-4=15-9x\)

\(< =>10x+9x=15+4\)

\(< =>19x=19< =>x=1\)

\(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)

\(< =>\frac{\left(10x+3\right).3}{36}=\frac{36}{36}+\frac{\left(6+8x\right).4}{36}\)

\(< =>\left(10x+3\right).3=36+\left(6+8x\right).4\)

\(< =>30x+9=36+24+32x\)

\(< =>32x-30x=9-36-24\)

\(< =>2x=9-60=-51< =>x=-\frac{51}{2}\)

\(\frac{4}{2x+3}-\frac{7}{3x-5}=0\left(đkxđ:x\ne-\frac{3}{2};\frac{5}{3}\right)\)

\(< =>\frac{4\left(3x-5\right)}{\left(2x+3\right)\left(3x-5\right)}-\frac{7\left(2x+3\right)}{\left(2x+3\right)\left(3x-5\right)}=0\)

\(< =>12x-20-14x-21=0\)

\(< =>2x+41=0< =>x=-\frac{41}{2}\left(tm\right)\)

\(\frac{4}{2x-3}+\frac{4x}{4x^2-9}=\frac{1}{2x+3}\left(đk:x\ne-\frac{3}{2};\frac{3}{2}\right)\)

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

\(< =>8x+12+4x-2x+3=0\)

\(< =>10x=15< =>x=\frac{15}{10}=\frac{3}{2}\left(ktm\right)\)

\(\frac{2}{2x+1}+\frac{x}{4x^2-1}=\frac{7}{2x-1}\left(đkxđ:x\ne-\frac{1}{2};\frac{1}{2}\right)\)

\(< =>\frac{2\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}+\frac{x}{\left(2x+1\right)\left(2x-1\right)}=\frac{7\left(2x+1\right)}{\left(2x+1\right)\left(2x-1\right)}\)

\(< =>4x-2+x=14x+7\)

\(< =>14x-5x=-2-7\)

\(< =>9x=-9< =>x=-\frac{9}{9}=-1\left(tm\right)\)

a/ ĐKXĐ: ...

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

\(\Leftrightarrow\frac{x+7}{x^2-1}=\frac{1}{2}\Leftrightarrow2x+14=x^2-1\)

\(\Leftrightarrow x^2-2x-15=0\Rightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)

b/ ĐKXĐ: ...

Đặt \(\frac{x-1}{x}=a\)

\(a-\frac{3}{2a}=-\frac{5}{2}\Leftrightarrow2a^2+5a-3=0\)

\(\Rightarrow\left[{}\begin{matrix}a=-3\\a=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\frac{x-1}{x}=-3\\\frac{x-1}{x}=\frac{1}{2}\end{matrix}\right.\)

c/ ĐKXĐ: ...

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

\(\Leftrightarrow2x^2+5x+9=2x^2+x-5\)

\(\Leftrightarrow4x=-14\Rightarrow x=-\frac{7}{2}\)