1: =>x+1/2=0 hoặc 2/3-2x=0

=>x=-1/2 hoặc x=1/3

2: =>7/6x=5/2:3,75=2/3

=>x=2/3:7/6=2/3*6/7=12/21=4/7

3: =>2x-3=0 hoặc 6-2x=0

=>x=3 hoặc x=3/2

4: =>-5x-1-1/2x+1/3=3/2x-5/6

=>-11/2x-3/2x=-5/6-1/3+1

=>-7x=-1/6

=>x=1/42

tìm x biết:

(3x-1) [- 1/2x+5]=0

1/4+1/3:(2x-1)=-5

[2x+3/5]² - 9/25=0

-5(x+1/5)-1/2(x-2/3)=3/2x - 5 /6

[x+1/2]x [2/3-2x]=0

17/2-|2x-3/4|=-7/4

2/3x-1/2x =5/12

(x+1/5)²+17/25=26/25

[x.44/7+3/7].11/5-3/7=-2

3[3x-1/2]+1/9=0

Toán lớp 6Tìm x

 Trả lời  Câu hỏi tương tự

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3(3x-1/2)^3+1/9=0

g: Ta có: \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)

\(\Leftrightarrow3\left(6x^2-5x+1\right)-\left(18x^2-29x+3\right)=0\)

\(\Leftrightarrow18x^2-15x+3-18x^2+29x-3=0\)

\(\Leftrightarrow14x=0\)

hay x=0

w

1. \(\left(x-4\right)^2-25=0\)

<=> (x-4+5).(x-4-5) = 0

<=> (x+1)(x-9) = 0

<=> \(\left[\begin{matrix}x+1=0\\x-9=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=-1\\x=9\end{matrix}\right.\)

Vậy phương trình có tập nghiệm S = {-1;9}

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

<=> (2x-1)(2x-1+2-x) = 0

<=> (2x-1)(x+1) = 0

<=> \(\left[\begin{matrix}2x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}2x=1\\x=-1\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=0.5\\x=-1\end{matrix}\right.\)

Vậy phương trình có tập nghiệm S = {-1 ; 0,5}

3. \(x^2+6x+9=4x^2\)

<=> \(\left(x+3\right)^2-4x^2=0\)

<=> (x+3+2x)(x+3-2x) = 0

<=> (3x+3)(3-x) = 0

<=> \(\left[\begin{matrix}3x+3=0\\3-x=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}3x=-3\\x=3\end{matrix}\right.\Leftrightarrow}\left[\begin{matrix}x=-1\\x=3\end{matrix}\right.\) Vậy phương trình có tập nghiệm S = {-1 ; 3}

4. (2x-5)(x+11) = (5-2x)(2x+1)

<=> (2x-5)(x+11) = - (2x-5)(2x+1)

<=> x + 11 = -2x - 1

<=> x+2x = -12

<=> 3x = -12

<=> x = -4

Vậy phương trình có một nghiệm duy nhất là x = -4

5. \(2x^2+5x+3=0\)

<=> \(2x^2+2x+3x+3=0\)

<=> \(2x\left(x+1\right)+3\left(x+1\right)=0\)

<=> \(\left(x+1\right)\left(2x+3\right)=0\)

<=> \(\left[\begin{matrix}x+1=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=-1\\2x=-3\end{matrix}\right.\Leftrightarrow}\left[\begin{matrix}x=-1\\x=\frac{-3}{2}\end{matrix}\right.\) Vậy phương trình có tập nghiệm S = { -1 ; -3/2 }

1) (x-4)^2-25=0

<=> (x-4+5)(x-4-5)=0

\(\Leftrightarrow\left[\begin{matrix}x=-1\\x=9\end{matrix}\right.\)

2) (2x-1)2+(2-x)(2x-1)=0

<=> (2x-1)(2+2-x)=0

<=> \(\left[\begin{matrix}x=\frac{1}{2}\\x=4\end{matrix}\right.\)

3) x^2+6x+9=4x^2

<=> 3x^2 -6x-9=0

<=> x^2 -2x -3=0

<=> x^2 -3x+x-3=0

<=> x(x-3)+(x-3)=0

<=> (x-3)(x+1)=0

=>\(\left[\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

4) (2x-5)(x+11)=(5-2x)(2x+1)

-(5-2x)(x+11)-(5-2x)(2x+1)=0

(5-2x)(x+11+2x+1)=0

=>\(\left[\begin{matrix}x=\frac{5}{2}\\x=-4\end{matrix}\right.\)

5)2x^2+5x+3=0

2x^2+2x+3x+3=0

2x(x+1)+3(x+1)=0

(x+1)(2x+3)=0

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

a) \(\left(x-4\right)^2-25=0\)

=> (x-4)² = 25

Suy ra \(\left\{\begin{matrix}x-4=5\\x-4=-5\end{matrix}\right.\)

=> \(\left\{\begin{matrix}x=5+4\\x=-5+4\end{matrix}\right.\)

=>\(\left\{\begin{matrix}x=9\\x=-1\end{matrix}\right.\)

2. (2x-1)² + (2-x)(2x-1)=0

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

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

\(\Rightarrow\left\{\begin{matrix}2x-1=0\\x+1=0\end{matrix}\right.\)

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

3. x²+6x+9=4x²

3x²-6x-9 =0

3(x² - 2x -3) =0

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

(x-3)(x+1) =0

\(\left\{\begin{matrix}x-3=0\\x+1=0\end{matrix}\right.\)

\(\left\{\begin{matrix}x=3\\x=-1\end{matrix}\right.\)

4. (2x-5)(x+11)=(5-2x)(2x+1)

(2x-5)(x+11) + (2x-5)(2x+1)=0

(2x-5)(3x+12) =0

\(\left\{\begin{matrix}2x-5=0\\3\left(x+4\right)=0\end{matrix}\right.\)

\(\left\{\begin{matrix}x=2,5\\x=-4\end{matrix}\right.\)

5. 2x²+5x+3=0

(2x²+2x)+(3x+3)=0

2x(x+1) +3(x+1) =0

(x+1)(2x+3) =0

\(\left\{\begin{matrix}x+1=0\\2x+3=0\end{matrix}\right.\)

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

a: (2x+1)(3-x)(4-2x)=0

=>(2x+1)(x-3)(x-2)=0

hay \(x\in\left\{-\dfrac{1}{2};3;2\right\}\)

b: 2x(x-3)+5(x-3)=0

=>(x-3)(2x+5)=0

=>x=3 hoặc x=-5/2

c: =>(x-2)(x+2)+(x-2)(2x-3)=0

=>(x-2)(x+2+2x-3)=0

=>(x-2)(3x-1)=0

=>x=2 hoặc x=1/3

d: =>(x-2)(x-3)=0

=>x=2 hoặc x=3

e: =>(2x+5+x+2)(2x+5-x-2)=0

=>(3x+7)(x+3)=0

=>x=-7/3 hoặc x=-3

f: \(\Leftrightarrow2x^3+5x^2-3x=0\)

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

\(\Leftrightarrow x\left(x+3\right)\left(2x-1\right)=0\)

hay \(x\in\left\{0;-3;\dfrac{1}{2}\right\}\)

1) (x+6)(3x-1)+x+6=0

⇔(x+6)(3x-1)+(x+6)=0

⇔(x+6)(3x-1+1)=0

⇔3x(x+6)=0

2) (x+4)(5x+9)-x-4=0

⇔(x+4)(5x+9)-(x+4)=0

⇔(x+4)(5x+9-1)=0

⇔(x+4)(5x+8)=0

3)(1-x)(5x+3)÷(3x-7)(x-1)

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

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bạn ơi !!!

đăng từng câu thôi thế này nhìn loạn cả mắt luôn á

rối mắt quá

a, 2/3+1/3:x=-7

1/3:x=-7-2/3

1/3:x=-23/3

x=1/3:-23/3

x=-1/23

Vậy x=-1/23

c, (2x-3)(6-2x)=0

*TH1: 2x-3=0

2x=3

x=3/2

*TH2: 6-2x=0

2x=6

x=6/2

x=3

Vậy x=3/2 hoặc x=3

d,x:3/4+1/4=-2/3

x:3/4=-2/3-1/4

x:3/4=-11/12

x=-11/12*3/4

x=-11/16