\(a,\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\) (x khác -3; khác 0)

\(=\frac{3}{2\left(x+3\right)}-\frac{x-6}{2x.\left(x+3\right)}=\frac{3x}{2x.\left(x+3\right)}-\frac{x-6}{2x.\left(x+3\right)}=\frac{3x-x+6}{2x.\left(x+3\right)}=\frac{2x+6}{x.\left(2x+6\right)}=\frac{1}{x}\)

\(b,\left(\frac{2x+1}{2x-1}-\frac{2x-1}{2x+1}\right):\frac{4x}{10x-5}\) (x khác 0 , khác 1/2 khác -1/2 )

\(=\left(\frac{\left(2x+1\right)^2}{\left(2x-1\right)\left(2x+1\right)}-\frac{\left(2x-1\right)^2}{\left(2x-1\right)\left(2x+1\right)}\right).\frac{10x-5}{4x}\)

\(=\left(\frac{4x^2+4x+1}{\left(2x-1\right)\left(2x+1\right)}-\frac{4x^2-4x+1}{\left(2x-1\right)\left(2x+1\right)}\right).\frac{10x-5}{4x}\)

\(=\frac{8x}{\left(2x-1\right)\left(2x+1\right)}.\frac{5.\left(2x-1\right)}{4x}=\frac{10}{2x+1}\)

\(c,\frac{x^2+x}{5x^2-10x+5}:\frac{3x+3}{5x-5}\) (x khác 1 ; khác -1)

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

\(\frac{x+5}{4}-\frac{2x-3}{3}=\frac{6x-1}{8}+\frac{2x-1}{12}\)

=> \(\frac{x+5}{4}-\frac{2x-3}{3}-\frac{6x-1}{8}-\frac{2x-1}{12}=0\)

=> \(\frac{6x+30}{24}-\frac{16x-24}{24}-\frac{18x-3}{24}-\frac{4x-2}{24}=0\)

=> \(\left(6x+30\right)-\left(16x-24\right)-\left(18x-3\right)-\left(4x-2\right)=0\)

=> \(6x+30-16x+24-18x+3-4x+2=0\)

=> \(\left(6-16-18-4\right)x+\left(30+24+3+2\right)=0\)

=> \(-32x+59=0\)

=> \(-32x=-59\)

=> \(x=\frac{-59}{-32}=\frac{59}{32}\)

\(\frac{x+5}{4}-\frac{2x-3}{3}=\frac{6x-1}{8}+\frac{2x-1}{12}\)

\(< =>\frac{6x+30}{24}-\frac{16x-24}{24}=\frac{18x-3}{24}+\frac{4x-2}{24}\)

\(< =>\frac{6x+30}{24}-\frac{16x-24}{24}-\frac{18x-3}{24}-\frac{4x-2}{24}=0\)

\(< =>6x+30-16x+24-18x+3-4x+2=0\)

\(< =>6x-16x-18x-4x+\left(30+24+3+2\right)=0\)

\(< =>x\left(6-16-18-4\right)+59=0\)

\(< =>x.\left(-32\right)=-59\)\(\)

\(< =>x=\frac{59}{32}\)

\(ĐKXĐ:x\ne-1;1\)

\(\frac{2x-6}{x+1}+\frac{x}{1-x}-\frac{6\left(x+1\right)}{x^2-1}=0\)

\(< =>\frac{2x-6}{x+1}+\frac{x}{1-x}-\frac{6x+6}{x^2-1}=0\)

\(< =>\frac{\left(2x-6\right)\left(1-x\right)+x^2+x}{\left(x+1\right)\left(1-x\right)}=\frac{6x+6}{x^2-1}\)

\(< =>\frac{2x-2x^2-6+6x+x^2+x}{x-x^2+1-x}=\frac{6x+6}{x^2-1}\)

\(< =>\frac{9x-x^2-6}{-x^2+1}=\frac{6x+6}{x^2-1}\)

\(< =>\frac{9x-x^2-6}{-x^2+1}=\frac{-6x-6}{-x^2+1}\)

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

\(< =>-x^2+12x=0\)\(< =>x^2-12x=0\)

\(< =>x\left(x-12\right)=0\)

\(< =>\orbr{\begin{cases}x=0\left(tmđk\right)\\x=12\left(tmđk\right)\end{cases}}\)

Vậy tập nghiệm pt trên là {0;12}

a) ta có: \(\frac{x-3}{6}=\frac{3}{2x-6}\)

\(\Rightarrow\left(x-3\right).\left(2x-6\right)=6.3\)

\(\Rightarrow x.\left(2x-6\right)-3.\left(2x-6\right)=18\)

\(2x^2-6x-6x+18=18\)

\(2x^2-12x+18=18\)

\(2x^2-12x=0\)

\(2x.\left(x-6\right)=0\)

\(\Rightarrow2x=0\Rightarrow x=0\)

\(x-6=0\Rightarrow x=6\)

KL: x =0 hoặc x = 6

b) ta có: \(\frac{x+6}{x+2}=\frac{2x+3}{2x-1}\)

\(\Rightarrow\left(x+6\right).\left(2x-1\right)=\left(x+2\right).\left(2x+3\right)\)

\(\Rightarrow x.\left(2x-1\right)+6.\left(2x-1\right)=x.\left(2x+3\right)+2.\left(2x+3\right)\)

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

\(2x^2+11x-6=2x^2+7x+6\)

\(\Rightarrow2x^2+11x-2x^2-7x=6+6\)

\(3x=12\)

\(x=12:3\)

\(x=4\)

\(\frac{x-3}{6}=\frac{3}{2x-6}\Leftrightarrow\left(x-3\right).\left(2x-6\right)=18\Leftrightarrow2x^2-12x+18=18\Leftrightarrow2x^2-12x=0\)

\(\Leftrightarrow x^2-6x=0\Leftrightarrow x.\left(x-6\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=6\end{cases}}\)

\(\frac{x+6}{x+2}=\frac{2x+3}{2x-1}\Leftrightarrow\left(x+6\right).\left(2x-1\right)=\left(2x+3\right).\left(x+2\right)\)

\(\Leftrightarrow2x^2+11x-6=2x^2+7x+6\Leftrightarrow4x-12=0\Leftrightarrow x=3\)

\(\frac{x-3}{6}=\frac{3}{2x-6}\)

\(\Rightarrow\left(x-3\right)\left(2x-6\right)=18\)

\(\Rightarrow2x^2-6x-6x+18=18\)

\(\Rightarrow2x^2-12x=0\)

\(\Rightarrow\left(2x-12\right).x=0\)

\(\Rightarrow\hept{\begin{cases}x=0\\2x-12=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x=6\end{cases}}}\)

a, \(\frac{x-3}{5}\) = 6 - \(\frac{1-2x}{3}\)

⇔ 3(x - 3) = 90 - 5(1 - 2x)

⇔ 3x - 9 = 90 - 5 + 10x

⇔ 3x - 10x = 90 - 5 + 9

⇔ -7x = 94

⇔ x = \(\frac{-94}{7}\)

S = { \(\frac{-94}{7}\) }

b, \(\frac{3x-2}{6}\) - 5 = \(\frac{3-2\left(x+7\right)}{4}\)

⇔ 2(3x - 2) - 60 = 9 - 6(x + 7)

⇔ 6x - 4 - 60 = 9 - 6x - 42

⇔ 6x + 6x = 9 - 42 + 60 + 4

⇔ 12x = 31

⇔ x = \(\frac{31}{12}\)

S = { \(\frac{31}{12}\) }

c, \(\frac{x+8}{6}\) - \(\frac{2x-5}{5}\) = \(\frac{x+1}{3}\) - x + 7

⇔ 5(x+ 8) - 6(2x - 5) = 10(x+1) - 30x+210

⇔ 5x+ 40 - 12x+ 30 = 10x+ 10 - 30x+210

⇔ 5x - 12x - 10x+ 30x = 10+ 210 - 30- 40

⇔ 13x = 150

⇔ x = \(\frac{150}{13}\)

S = { \(\frac{150}{13}\) }

d, \(\frac{7x}{8}\) - 5(x - 9) = \(\frac{2x+1,5}{6}\)

⇔ 21x - 120(x - 9) = 4(2x + 1,5)

⇔ 21x - 120x + 1080 = 8x + 6

⇔ 21x - 120x - 8x = 6 - 1080

⇔ -107x = -1074

⇔ x = \(\frac{1074}{107}\)

S = { \(\frac{1074}{107}\) }

e, \(\frac{5\left(x-1\right)+2}{6}\) - \(\frac{7x-1}{4}\) = \(\frac{2\left(2x+1\right)}{7}\) - 5

⇔ 140(x-1)+56 - 42(7x-1) = 48(2x+1)-840

⇔ 140x -140+56 -294x+42= 96x+48 -840

⇔ 140x -294x -96x = 48 -840 -42 -56+140

⇔ -250x = -750

⇔ x = 3

S = { 3 }

f, \(\frac{x+1}{3}\) + \(\frac{3\left(2x+1\right)}{4}\) = \(\frac{2x+3\left(x+1\right)}{6}\) + \(\frac{7+12x}{12}\)

⇔ 4(x+1)+9(2x+1) = 4x+6(x+1)+7+12x

⇔ 4x+4+18x+9 = 4x+6x+6+7+12x

⇔ 4x+18x - 4x - 6x - 12x = 6+7- 9 - 4

⇔ 0x = 0

S = R

Chúc bạn học tốt !

Vì số lượng bài khá nhiều và mình cũng không có quá nhiều thời gian nên không tránh khỏi sai sót, nếu phát hiện mong bạn thông cảm! Bài của tớ làm khá tắt bước, chỉ nên tham khảo. Bạn có thể tự biểu diễn tập nghiệm được không?

a. \(x+8>3x-1\)

\(\Leftrightarrow-2x>-9\)

\(\Leftrightarrow x< \frac{9}{2}\)

b. \(3x-\left(2x+5\right)\le\left(2x-3\right)\)

\(\Leftrightarrow3x-2x-5\le2x-3\)

\(\Leftrightarrow-x\le2\)

\(\Leftrightarrow x\ge2\)

c. \(\left(x-3\right)\left(x+3\right)< x\left(x+2\right)+3\)

\(\Leftrightarrow x^2-9< x^2+2x+3\)

\(\Leftrightarrow2x>-12\Leftrightarrow x>-6\)

d. \(2\left(3x-1\right)-2x< 2x+1\)

\(\Leftrightarrow6x-2-2x< 2x+1\)

\(\Leftrightarrow2x< 3\)

\(\Leftrightarrow x< \frac{3}{2}\)

e. \(\frac{3-2x}{5}>\frac{2-x}{3}\)

\(\Leftrightarrow3\left(3-2x\right)>5\left(2-x\right)\)

\(\Leftrightarrow9-6x>10-5x\)

\(\Leftrightarrow-x>1\) \(\Leftrightarrow x< -1\)

f. \(\frac{x-2}{6}-\frac{x-1}{3}\le\frac{x}{2}\)

\(\Leftrightarrow x-2-2\left(x-1\right)\le3x\)

\(\Leftrightarrow x-2-2x+2\le3x\)

\(\Leftrightarrow-4x\le0\Leftrightarrow x\ge0\)

g. \(\frac{x+1}{3}>\frac{2x-1}{6}\ge4\)

\(\Leftrightarrow2x+2>2x-1\ge24\)

\(\Leftrightarrow2x+2>2x\ge25\)

\(\Leftrightarrow x\ge\frac{25}{2}\)

h. \(1+\frac{2x+1}{3}>\frac{2x-1}{6}-2\)

\(\Leftrightarrow6+4x+2>2x-1-12\)

\(\Leftrightarrow2x>-25\)

\(\Leftrightarrow x>-\frac{25}{2}\)

i. \(\frac{x+5}{6}-\frac{2x+1}{3}\le\frac{x+3}{2}\)

\(\Leftrightarrow x+5-4x-2\le3x+9\)

\(\Leftrightarrow-6x\le6\)

\(\Leftrightarrow x\ge-1\)

j. \(\frac{5x+4}{6}-\frac{2x-1}{12}\ge4\)

\(\Leftrightarrow10x+8-2x+1\ge48\)

\(\Leftrightarrow8x\ge39\)

\(\Leftrightarrow x\ge\frac{39}{8}\)

Bạn tự biểu diễn nghiệm trên trục số nhé!

a) \(x+8>3x-1\)

\(\Leftrightarrow x-3x>-8-1\)

\(\Leftrightarrow-2x>-9\)

\(\Leftrightarrow x< \frac{9}{2}\)

b)

d)

e) \(\frac{3-2x}{5}>\frac{2-x}{3}\)

\(\Leftrightarrow\frac{\left(3-2x\right)\times3}{15}>\frac{\left(2-x\right)\times5}{15}\)

\(\Leftrightarrow9-6x>10-5x\)

\(\Leftrightarrow-6x+5x>-9+10\)

\(\Leftrightarrow-x>1\)

\(\Leftrightarrow x< -1\)

f)\(\frac{x-2}{6}-\frac{x-1}{3}\le\frac{x}{2}\)

\(\Leftrightarrow x-2-2\left(x-1\right)\le3x\)

\(\Leftrightarrow x-2-2x+2\le3x\)

\(\Leftrightarrow-4x\le0\)

\(\Leftrightarrow x\ge0\)

g) \(\frac{x+1}{3}>\frac{2x-1}{6}\ge4\)

\(\Leftrightarrow\frac{\left(x+1\right)\cdot2}{6}>\frac{2x-1}{6}\ge\frac{4\cdot6}{6}\)

\(\Leftrightarrow2x+2>2x+1\ge24\)

\(\Leftrightarrow2x+2>2x\ge25\)

\(\Leftrightarrow x\ge\frac{25}{2}\)

h)\(1+\frac{2x+1}{3}>\frac{2x-1}{6}-2\)

\(\Leftrightarrow\frac{1}{6}+\frac{\left(2x+1\right)\cdot2}{6}>\frac{2x-1}{6}-\frac{2\cdot6}{6}\)

\(\Leftrightarrow6+4x+2>2x-1-12\)

\(\Leftrightarrow2x>-21\)

\(\Leftrightarrow x>\frac{-21}{2}\)

i)\(\frac{x+5}{6}-\frac{2x+1}{3}\le\frac{x+3}{2}\)

\(\Leftrightarrow\frac{x+5}{6}-\frac{\left(2x+1\right)\cdot2}{6}\le\frac{\left(x+3\right)\cdot3}{6}\)

\(\Leftrightarrow x+5-4x+2\le3x+9\)

\(\Leftrightarrow-3x-x+4x\le9-5-2\)

\(\Leftrightarrow x\le2\)

j) \(\frac{5x+4}{6}-\frac{2x-1}{12}\ge4\)

\(\Leftrightarrow\frac{\left(5x+4\right)\cdot2}{12}-\frac{2x-1}{12}\ge\frac{4\cdot12}{12}\)

\(\Leftrightarrow10x+8-2x-1\ge48\)

\(\Leftrightarrow10x-2x\ge48-8+1\)

\(\Leftrightarrow8x\ge41\)

\(\Leftrightarrow x\ge\frac{41}{8}\)

Mình không chắc là mình làm đúng đâu. Nhưng có sai sót gì thì cứ nói cho mình biết. Chúc bạn học tốt ^-^

a, »x – 3x > -1 – 8

»-2x > -9

» x < \(\frac{9}{2}\)

b, » 3x – 2x – 5 ≤ 2x – 3

» 3x – 2x – 2x ≤ -3+5

» -x ≤ 2

» x ≥ -2

c, » x² + 3 x - 3 x - 9 < x² + 2x + 3

» x² + 3 x – 3 x – x² -2 x < 3 + 9

» -2x < 12

» x > -6

d, » 6 x -2 - 2 x < 2x + 1

» 6 x – 2 x – 2 x < 1+2

» 2x < 3

» x < \(\frac{3}{2}\)

Hướng dẫn:

a) Đặt : \(x^2-2x+1=t\)Ta có: 

\(\frac{1}{t+1}+\frac{2}{t+2}=\frac{6}{t+3}\)

b) Đặt : \(x^2+2x+1=t\)

Ta có pt: \(\frac{t}{t+1}+\frac{t+1}{t+2}=\frac{7}{6}\)

c)ĐK: x khác 0

Đặt: \(x+\frac{1}{x}=t\)

KHi đó: \(x^2+\frac{1}{x^2}=t^2-2\)

Ta có pt: \(t^2-2-\frac{9}{2}t+7=0\)

a) Đặt \(x^2-2x+3=v\)

Phương trình trở thành \(\frac{1}{v-1}+\frac{2}{v}=\frac{6}{v+1}\)

\(\Rightarrow\frac{v\left(v+1\right)+2\left(v+1\right)\left(v-1\right)}{v\left(v+1\right)\left(v-1\right)}=\frac{6v\left(v-1\right)}{v\left(v+1\right)\left(v-1\right)}\)

\(\Rightarrow v\left(v+1\right)+2\left(v+1\right)\left(v-1\right)=6v\left(v-1\right)\)

\(\Rightarrow v^2+v+2v^2-2=6v^2-6v\)

\(\Rightarrow3v^2-7v+2=0\)

Ta có \(\Delta=7^2-4.3.2=25,\sqrt{\Delta}=5\)

\(\Rightarrow\orbr{\begin{cases}v=\frac{7+5}{6}=2\\v=\frac{7-5}{6}=\frac{1}{3}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x^2-2x+3=2\\x^2-2x+3=\frac{1}{3}\end{cases}}\)

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

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

\(\Rightarrow x^2-2x+\frac{8}{3}=0\)

Ta có \(\Delta=2^2-4.\frac{8}{3}=\frac{-20}{3}< 0\)

Vậy phương trình có 1 nghiệm là x = 1

c) Đặt \(\left(x+\frac{1}{x}\right)=a\) Khi đó pt có dạng :

\(a^2-\frac{9}{2}a+7-2=0\)

\(\Leftrightarrow2a^2-9a+10=0\)

\(\Leftrightarrow2a^2-4a-5a+10=0\)

\(\Leftrightarrow\left(a-2\right)\left(2a-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=2\\a=\frac{5}{2}\end{cases}}\)

+) Với \(a=\frac{5}{2}\Rightarrow x+\frac{1}{x}=\frac{5}{2}\)

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

\(\Rightarrow2x^2+2-5x=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{1}{2}\end{cases}}\) ( thỏa mãn)

+) Với \(a=2\Rightarrow x+\frac{1}{x}=2\)

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

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

\(\Leftrightarrow x=1\) ( thỏa mãn )

Vậy pt đã cho có tập nghiệm \(S=\left\{1,\frac{1}{2},2\right\}\)