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15 tháng 4 2020

Đây là lớp 8 nha các b giúp mk với

Do mk viết nhầm

Bài 1:

a) Ta có: \(\frac{4}{5}x-3=\frac{1}{5}x\left(4x-15\right)\)

\(\Leftrightarrow\frac{4x}{5}-3=\frac{4x^2}{5}-3x\)

\(\Leftrightarrow\frac{12x}{15}-\frac{45}{15}-\frac{12x^2}{15}+\frac{45x}{15}=0\)

Suy ra: \(12x-45-12x^2+45x=0\)

\(\Leftrightarrow-12x^2+57x-45=0\)

\(\Leftrightarrow-12x^2+12x+45x-45=0\)

\(\Leftrightarrow-12x\left(x-1\right)+45\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(-12x+45\right)=0\)

\(\Leftrightarrow-3\left(x-1\right)\left(4x-15\right)=0\)

\(-3\ne0\)

nên \(\left[{}\begin{matrix}x-1=0\\4x-15=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\4x=15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{15}{4}\end{matrix}\right.\)

Vậy: Tập nghiệm \(S=\left\{1;\frac{15}{4}\right\}\)

b) Ta có: \(\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}=\frac{\left(x-3\right)\left(3-x\right)}{4}\)

\(\Leftrightarrow\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}+\frac{\left(x-3\right)^2}{4}=0\)

\(\Leftrightarrow\frac{12\left(x-3\right)}{12}-\frac{2\left(x-3\right)\left(2x-5\right)}{12}+\frac{3\left(x-3\right)^2}{12}=0\)

Suy ra: \(12\left(x-3\right)-2\left(2x^2-11x+15\right)+3\left(x^2-6x+9\right)=0\)

\(\Leftrightarrow12x-36-4x^2+22x-30+3x^2-18x+27=0\)

\(\Leftrightarrow-x^2+16x-39=0\)

\(\Leftrightarrow-\left(x^2-16x+39\right)=0\)

\(\Leftrightarrow x^2-13x-3x+39=0\)

\(\Leftrightarrow x\left(x-13\right)-3\left(x-13\right)=0\)

\(\Leftrightarrow\left(x-13\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-13=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=13\\x=3\end{matrix}\right.\)

Vậy: Tập nghiệm S={3;13}

c) Ta có: \(\frac{\left(3x+1\right)\left(3x-2\right)}{3}+5\left(3x+1\right)=\frac{2\left(2x+1\right)\left(3x+1\right)}{3}+2x\left(3x+1\right)\)

\(\Leftrightarrow\frac{9x^2-3x-2}{3}+5\left(3x+1\right)-\frac{12x^2+10x+2}{3}-2x\left(3x+1\right)=0\)

\(\Leftrightarrow\frac{9x^2-3x-2-12x^2-10x-2}{3}-6x^2+13x+5=0\)

\(\Leftrightarrow\frac{-3x^2-13x-4}{3}+\frac{3\left(-6x^2+13x+5\right)}{3}=0\)

Suy ra: \(-3x^2-13x-4-18x^2+39x+15=0\)

\(\Leftrightarrow-21x^2+26x+11=0\)

\(\Leftrightarrow-21x^2-7x+33x+11=0\)

\(\Leftrightarrow-7x\left(3x+1\right)+11\left(3x+1\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(-7x+11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\-7x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\-7x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{3}\\x=\frac{11}{7}\end{matrix}\right.\)

Vậy: Tập nghiệm \(S=\left\{-\frac{1}{3};\frac{11}{7}\right\}\)

8 tháng 1 2020

1.

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

\(MC:12\)

Quy đồng :

\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)

\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)

\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)

\(\Leftrightarrow6x+9-3x=-4-9+16\)

\(\Leftrightarrow-7x=3\)

\(\Leftrightarrow x=\frac{-3}{7}\)

2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)

\(MC:20\)

Quy đồng :

\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)

\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)

\(\Leftrightarrow30x+15-20=15x-2\)

\(\Leftrightarrow15x=3\)

\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)

26 tháng 6 2018

a) Qui đồng rồi khử mẫu ta được:

   3(3x+2)-(3x+1)=2x.6+5.2

<=> 9x+6-3x-1 = 12x+10

<=> 9x-3x-12x  = 10-6+1

<=> -6x            = 5

<=> x               = -5/6

Vậy ....

b) ĐKXĐ: \(x\ne\pm2\)

Qui đồng rồi khử mẫu ta được:

   (x+1)(x+2)+(x-1)(x-2) = 2(x2+2)

<=> x2+3x+2+x2-3x+2 = 2x2+4

<=> x2+x2-2x2+3x-3x = 4-2-2

<=> 0x             = 0

<=> x vô số nghiệm

Vậy x vô số nghiệm với x khác 2 và x khác -2

c) \(\left(2x+3\right)\left(\frac{3x+7}{2-7x}+1\right)=\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)\) (ĐKXĐ:x khắc 2/7)

\(\Leftrightarrow\left(2x+3\right)\left(\frac{3x+8}{2-7x}+1\right)-\left(x-5\right)\left(\frac{3x+8}{2-7x}+1\right)=0\)

\(\Leftrightarrow\left(\frac{3x+8}{2-7x}+1\right)\left[\left(2x+3\right)-\left(x-5\right)\right]=0\)

\(\Leftrightarrow\left(\frac{3x+8}{2-7x}+1\right)\left(x+8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{3x+8}{2-7x}+1=0\\x+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{3x+8}{2-7x}=-1\\x+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}3x+8=-1\left(2-7x\right)\\x=0-8\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}3x+8=-2+7x\\x=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}-4x=-10\\x=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{5}{2}\\x=-8\end{cases}}}\) (nhận)

Vậy ...... 

d) (x+1)2-4(x2-2x+1) = 0

<=> x2+2x+1-4x2+8x-4 = 0

<=> -3x2+10x-3 = 0

giải phương trình

11 tháng 3 2020

1) \(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\)

<=> \(\frac{21x}{24}-\frac{100\left(x-9\right)}{24}=\frac{80x+6}{24}\)

<=> 21x - 100x + 900 = 80x + 6

<=> -79x - 80x = 6 - 900

<=> -159x = -894

<=> x = 258/53

Vậy S = {258/53}

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

<=> \(\frac{3\left(4x^2+4x+1\right)}{15}-\frac{5\left(x^2+2x+1\right)}{15}=\frac{7x^2-14x-5}{15}\)

<=> 12x2 + 12x + 3 - 5x2 - 10x - 5 = 7x2 - 14x - 5

<=> 7x2 + 2x - 7x2 + 14x = -5 + 2

<=> 16x = 3

<=> x = 3/16

Vậy S  = {3/16}

11 tháng 3 2020

3) 4(3x - 2) - 3(x - 4) = 7x+  10

<=> 12x - 8 - 3x + 12 = 7x + 10

<=> 9x - 7x = 10 - 4

<=> 2x = 6

<=> x = 3

Vậy S = {3}

4) \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)

<=> \(\frac{x^2+14x+40}{12}+\frac{3\left(x^2+2x-8\right)}{12}=\frac{4\left(x^2+8x-20\right)}{12}\)

<=> x2 + 14x + 40 + 3x2 + 6x - 24 = 4x2 + 32x - 80

<=> 4x2 + 20x - 4x2 - 32x = -80 - 16

<=> -12x = -96

<=> x = 8

Vậy S = {8}

\(\Leftrightarrow20\left(x^2-4x+3\right)-24\left(4x^2-4x+1\right)=15\left(9x^2+6x+1\right)+90x\left(x-1\right)\)

\(\Leftrightarrow20x^2-80x+60-96x^2+96x-24=135x^2+90x+15+90x^2-90x\)

\(\Leftrightarrow-301x^2+16x+21=0\)

\(\text{Δ}=16^2-4\cdot\left(-301\right)\cdot21=25540\)

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

\(\left\{{}\begin{matrix}x_1=\dfrac{-16-\sqrt{25540}}{-602}=\dfrac{16+\sqrt{25540}}{602}\\x_2=\dfrac{16-\sqrt{25540}}{602}\end{matrix}\right.\)

20 tháng 1 2019

a) <=> \(6x^2-5x+3-2x+3x\left(3-2x\right)=0\)

<=> \(6x^2-5x+3-2x+9x-6x^2=0\)

<=> \(2x+3=0\)

<=> \(x=\frac{-3}{2}\)

b) <=> \(10\left(x-4\right)-2\left(3+2x\right)=20x+4\left(1-x\right)\)

<=> \(10x-40-6-4x=20x+4-4x\)

<=> \(6x-46-16x-4=0\)

<=> \(-10x-50=0\)

<=> \(-10\left(x+5\right)=0\)

<=> \(x+5=0\)

<=> \(x=-5\)

c) <=> \(8x+3\left(3x-5\right)=18\left(2x-1\right)-14\)

<=> \(8x+9x-15=36x-18-14\)

<=> \(8x+9x-36x=+15-18-14\)

<=> \(-19x=-14\)

<=> \(x=\frac{14}{19}\)

d) <=>\(2\left(6x+5\right)-10x-3=8x+2\left(2x+1\right)\)

<=> \(12x+10-10x-3=8x+4x+2\)

<=> \(2x-7=12x+2\)

<=> \(2x-12x=7+2\)

<=> \(-10x=9\)

<=> \(x=\frac{-9}{10}\)

e) <=> \(x^2-16-6x+4=\left(x-4\right)^2\)

<=> \(x^2-6x-12-\left(x-4^2\right)=0\)

<=> \(x^2-6x-12-\left(x^2-8x+16\right)=0\)

<=> \(x^2-6x-12-x^2+8x-16=0\)

<=> \(2x-28=0\)

<=> \(2\left(x-14\right)=0\)

<=> x-14=0

<=> x=14

20 tháng 1 2019

Luffy , cậu sai câu c nhé , kia là -17 ạ => x=17/19

24 tháng 1 2020

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

\(\Leftrightarrow6x-2-5-3x=6x-3\)

\(\Leftrightarrow6x-3x-6x=-3+2+5\)

\(\Leftrightarrow-3x=4\)

\(\Leftrightarrow x=-\frac{4}{3}\)

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

\(\Leftrightarrow3x-\frac{3}{2}+\frac{4}{3}x-\frac{4}{3}=\frac{x}{4}\)

\(\Leftrightarrow3x+\frac{4}{3}x-\frac{x}{4}=\frac{3}{2}+\frac{4}{3}\)

\(\Leftrightarrow\frac{49}{12}x=\frac{17}{6}\)

\(\Leftrightarrow x=\frac{34}{49}\)

c) \(\frac{1}{5}\left(x-\frac{1}{3}\right)-4\left(\frac{x}{5}-\frac{1}{2}\right)=x\)

\(\Leftrightarrow\frac{1}{5}x-\frac{1}{15}-\frac{4}{5}x+2=x\)

\(\Leftrightarrow\frac{1}{5}x-\frac{4}{5}x-x=\frac{1}{15}-2\)

\(\Leftrightarrow-\frac{8}{5}x=-\frac{29}{15}\)

\(\Leftrightarrow x=\frac{29}{24}\)