PP chung ở cả 3 câu,nói ngắn gọn nhé:

Chứng mình x khác 0,hay nói cách khác x=0 không là nghiệm của phương trình.

Chia cả tử và mẫu cho x ,rồi giải bình thường bằng cách đặt ẩn phụ.

Vd ở câu a>>>4/(4x-8+7/x)+3/(4x-10+7/x)=1.Sau đó đặt 4x+7/x=a>>>4/(a-8)+3/(a-10)=1>>>giải bình thường,các câu sau tương tự

\(\frac{3x-7}{5}=\frac{2x-1}{3}\)

\(\Leftrightarrow9x-21=10x-5\)

\(\Leftrightarrow-x=16\Leftrightarrow x=-16\)

\(\frac{4x-7}{12}-x=\frac{3x}{8}\)

\(\Leftrightarrow\frac{4x-7-12x}{12}=\frac{3x}{8}\)

\(\Leftrightarrow\frac{-7-8x}{12}=\frac{3x}{8}\)

\(\Leftrightarrow-56-64x=36x\)

\(\Leftrightarrow-56=100x\Leftrightarrow x=\frac{-14}{25}\)

\(\frac{x-2009}{1234}+\frac{x-2009}{5678}-\frac{x-2009}{197}=0\)

\(\Leftrightarrow\left(x-2019\right)\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)=0\)

Vì \(\left(\frac{1}{1234}+\frac{1}{5678}-\frac{1}{197}\right)\ne0\)nên x - 2019 = 0

Vậy x = 2019

\(\frac{5x-8}{3}=\frac{1-3x}{2}\)

\(\Leftrightarrow10x-16=3-9x\)

\(\Leftrightarrow19x=19\Leftrightarrow x=1\)

\(\frac{x-5}{6}-\frac{x-9}{4}=\frac{5x-3}{8}+2\)

\(\Rightarrow\frac{4x-20-6x+54}{24}=\frac{5x-3+16}{8}\)

\(\Rightarrow\frac{-2x+34}{24}=\frac{5x+13}{8}\)

\(\Rightarrow-16x-272=120x+312\)

\(\Leftrightarrow-136x=584\Leftrightarrow x=\frac{-73}{17}\)

Ta có:

\(5\left(x^3-9x\right)=5x^3-45x.\)(1)

\(\left(15-5x\right).\left(-x^2-3x\right)=-15x^2-45x+5x^3+15x^2=5x^3-45x\)(2)

Từ (1)(2) suy ra \(5\left(x^3-9x\right)=\left(15-5x\right)\left(-x-3x\right)\)

\(\Rightarrow\frac{x^3-9x}{15-5x}=\frac{-x^2-3x}{5}\)(Điều phải chứng minh)

(8x - 3)(3x + 2) - (4x + 7)(x + 4) = (2x + 1)(5x - 1) - 33

=> 24x² + 16x - 9x - 6 - 4x² - 16x - 7x - 28 = 10x² - 2x + 5x - 1 - 33

=> 10x² - 19x = 0

=> x(10x - 19) = 0

=> x = 0

hoặc 10x - 19 = 0 => x = 19/10

                                                               Vậy x = 0, x = 19/10

TA CÓ: \(\left|x+1\right|=\frac{1}{2}\)

TH1: \(x+1=\frac{1}{2}\)                                                                      TH2: \(x+1=\frac{-1}{2}\)

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

THAY X =-1 /2 VÀO BIỂU THỨC M                                                         THAY X = -3/2 VÀO BIỂU THỨC M

\(M=5.\left(\frac{-1}{2}\right)^2-7.\frac{-1}{2}+\frac{1}{3}.\frac{-1}{2}-1\)                           \(M=5.\left(\frac{-3}{2}\right)^2-7.\frac{-3}{2}+\frac{1}{3}.\frac{-3}{2}-1\)

\(M=5.\frac{1}{4}-\frac{-1}{2}.\left(\frac{1}{3}+7\right)-1\)                                            \(M=5.\frac{9}{4}-\frac{-3}{2}.\left(\frac{1}{3}+7\right)-1\)

\(M=\frac{5}{4}-\frac{-1}{2}.\frac{22}{3}-1\)                                                              \(M=\frac{45}{4}-\frac{-3}{2}.\frac{22}{3}-1\)

\(M=\frac{5}{4}-\frac{-11}{3}-1\)                                                                      \(M=\frac{45}{4}-\left(-11\right)-1\)

\(M=\frac{47}{12}\)                                                                                              \(M=\frac{85}{4}\)

VẬY BIỂU THỨC M = 47/12 TẠI X= -1/2

................................= 85/4..............-3/2

CHÚC BN HỌC TỐT!!!!!!

a, 4x=5y=> x/5=y/4 => x/5=y/4=3x/15=2y/8 => 3x-2y/15-8=35/7=5( theo tính chất dãy tỉ số bằng nhau) => x=25;y=20 b, x/2=y/3=z/5 =>x+y+z/2+3+5=-90/10=-9(theo tính chất dãy tỉ số bằng nhau) =>x=-18;y=-27;z=-45 c, x:y:z=3:5:(-2) => x/3=y/5=z/-2 =5x/15=y/5=3z/-6 =>5x-y+3z/15-5+(-6)(theo tính chất dãy tỉ số bằng nhau) =124/4=31 =>x=93;y=155;z=-62 Mik sẽ bổ sung sau vì máy mik sắp hết pin

Ta có : \(\left(5x-3\right)^2-\frac{1^2}{64}=0\)

\(\Leftrightarrow\left(5x-3\right)^2=\frac{1}{64}\)

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

\(\Leftrightarrow\orbr{\begin{cases}5x-3=\frac{1}{8}\\5x-3=-\frac{1}{8}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}5x=\frac{1}{8}+3\\5x=-\frac{1}{8}+3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}5x=\frac{25}{8}\\5x=\frac{23}{8}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{25}{8}.\frac{1}{5}\\x=\frac{23}{8}.\frac{1}{5}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{8}\\x=\frac{23}{40}\end{cases}}\)

b) 3x - 7.(5x-1) = 6 - 2.(4-3x)

=> 3x - 35x + 7 = 6 - 8 + 6x

=> 3x - 35x - 6x = 6-8 -7

-38x = -9

x = 9/38

\(\frac{1}{5}\)ở đâu ra vậy bạn

=>(3x+2)(5x+1)=(5x+7)(3x-1)

(3x)(5x+1)+2(5x+1)=(5x)(3x-1)+7(3x-1)

15x²+3x+10x+2=15x²-5x+21x-7

(15x²-15x²)+(3x+10x+5x-21x)=-7-2

0-3x=-9

-3x=-9

x=(-9)/(-3)

x=3

\(\frac{3x+2}{5x+7}\)=\(\frac{3x-1}{5x+1}\)

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

\(\Rightarrow\)\(\frac{3}{3x-1}\)= \(\frac{6}{5x+1}\)\(\Rightarrow\)3.(5x+1) = 6.(3x-1)

\(\Rightarrow\)15x+3 = 18x -6

\(\Rightarrow\)3x = 6 +3

\(\Rightarrow\)x =3

b) PT \(\Leftrightarrow15x\left(5x+3\right)-35\left(5x+3\right)=0\)

\(\Leftrightarrow\left(15x-35\right)\left(5x+3\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{3}{5}\end{matrix}\right.\)

 Vậy \(S=\left\{-\dfrac{3}{5};\dfrac{7}{3}\right\}\)

c) PT \(\Leftrightarrow\left(2-3x\right)\left(x-11\right)+\left(2-3x\right)\left(2-5x\right)=0\)

\(\Leftrightarrow\left(2-3x\right)\left(-9-4x\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{9}{4}\end{matrix}\right.\)

  Vậy \(S=\left\{\dfrac{2}{3};-\dfrac{9}{4}\right\}\)

 

a)(x-1)(5x+3)=(3x-8)(x-1)

\(\Leftrightarrow\)(x-1)(5x+3)-(3x-8)(x-1)=0

\(\Leftrightarrow\left(x-1\right)\left(5x-3-3x+8\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x-5\right)=0\)

\(\left[{}\begin{matrix}x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{5}{2}\end{matrix}\right.\)

Vậy \(x\in\left\{1;\dfrac{5}{2}\right\}\)

a) Ta có: \(\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)

\(\Leftrightarrow5x^2+3x-5x-3=3x^2-3x-8x+8\)

\(\Leftrightarrow5x^2-2x-3=3x^2-11x+8\)

\(\Leftrightarrow5x^2-2x-3-3x^2+11x-8=0\)

\(\Leftrightarrow2x^2+9x-11=0\)

\(\Leftrightarrow2x^2+11x-2x-11=0\)

\(\Leftrightarrow x\left(2x+11\right)-\left(2x+11\right)=0\)

\(\Leftrightarrow\left(2x+11\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+11=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-11\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{11}{2}\\x=1\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{11}{2};1\right\}\)

b) Ta có: \(3x\left(25x+15\right)-35\left(5x+3\right)=0\)

\(\Leftrightarrow3x\cdot5\cdot\left(5x+3\right)-35\left(5x+3\right)=0\)

\(\Leftrightarrow15x\left(5x+3\right)-35\left(5x+3\right)=0\)

\(\Leftrightarrow\left(5x+3\right)\left(15x-35\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+3=0\\15x-35=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-3\\15x=35\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=\dfrac{7}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{3}{5};\dfrac{7}{3}\right\}\)

c) Ta có: \(\left(2-3x\right)\left(x-11\right)=\left(3x-2\right)\left(2-5x\right)\)

\(\Leftrightarrow2x-22-3x^2+33x=6x-15x^2-4+10x\)

\(\Leftrightarrow-3x^2+35x-22=-15x^2+16x-4\)

\(\Leftrightarrow-3x^2+35x-22+15x^2-16x+4=0\)

\(\Leftrightarrow12x^2+19x-18=0\)

\(\Leftrightarrow12x^2+27x-8x-18=0\)

\(\Leftrightarrow3x\left(4x+9\right)-2\left(4x+9\right)=0\)

\(\Leftrightarrow\left(4x+9\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+9=0\\3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-9\\3x=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{9}{4}\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{9}{4};\dfrac{2}{3}\right\}\)