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3 tháng 11 2018

Ta có :

\(\dfrac{2x+3}{5x+2}=\dfrac{4x+5}{10x+2}\)

\(\Leftrightarrow\left(2x+3\right)\left(10x+2\right)=\left(4x+5\right)\left(5x+2\right)\)

\(\Leftrightarrow2x\left(10x+2\right)+3\left(10x+2\right)=4x\left(5x+2\right)+5\left(5x+2\right)\)

\(\Leftrightarrow20x^2+4x+30x+6=20x^2+8x+25x+10\)

\(\Leftrightarrow20x^2+34x+6=20x^2+33x+10\)

\(\Leftrightarrow20x^2+34x+6-20x^2-33x-10=0\)

\(\Leftrightarrow x-4=0\)

\(\Leftrightarrow x=4\)

Vậy x = 4

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

\(\Leftrightarrow3\left|x\right|=\dfrac{4}{5}-\dfrac{1}{5}=\dfrac{3}{5}\)

\(\Leftrightarrow\left|x\right|=\dfrac{1}{5}\)

hay \(x\in\left\{\dfrac{1}{5};-\dfrac{1}{5}\right\}\)

b) Ta có: \(4x-\dfrac{1}{2}x+\dfrac{3}{5}x=\dfrac{4}{5}\)

nên \(\dfrac{41}{10}x=\dfrac{4}{5}\)

hay \(x=\dfrac{8}{41}\)

c) Ta có: \(\left(2x-8\right)\left(10-5x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-8=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=8\\5x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

d) Ta có: \(\dfrac{3}{4}+\dfrac{1}{4}\left|2x-1\right|=\dfrac{7}{2}\)

\(\Leftrightarrow\dfrac{1}{4}\left|2x-1\right|=\dfrac{7}{2}-\dfrac{3}{4}=\dfrac{14}{4}-\dfrac{3}{4}=\dfrac{11}{4}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=11\\2x-1=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=12\\2x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-5\end{matrix}\right.\)

28 tháng 10 2017

\(\dfrac{2x+3}{5x+2}=\dfrac{4x+5}{10x+2}\\ \Leftrightarrow\dfrac{4x+6}{10x+4}=\dfrac{4x+5}{10x+2}\)

Áp dụng t.c của dãy tỉ số bằng nhau, ta có:

\(\dfrac{4x+6}{10x+4}=\dfrac{4x+5}{10x+2}=\dfrac{4x+6-4x-5}{10x+4-10x-2}=\dfrac{1}{2}\)

=>\(4x+6=\dfrac{1}{2}\left(10x+4\right)\)

=>4x+6=5x+2

=>x=6-2=4

Vậy x=4

28 tháng 10 2017

Mình sẽ làm cách dãy tỉ số bằng nhau ,vì nhân sẽ khá là rối =.=

\(\dfrac{2x+3}{5x+2}=\dfrac{2\left(2x+3\right)}{2\left(5x+2\right)}=\dfrac{4x+6}{10x+4}\)

Hay \(\dfrac{4x+6}{10x+4}=\dfrac{4x+5}{10x+2}=\dfrac{4x+6-4x-5}{10x+4-10x-2}=\dfrac{1}{2}\)

Thay vào ta có:

\(\dfrac{2x+3}{5x+2}=\dfrac{1}{2}\Leftrightarrow5x+2=4x+6\Leftrightarrow5x=4x+4\Leftrightarrow x=4\)

`@` `\text {Ans}`

`\downarrow`

`a)`

`3x(4x-1) - 2x(6x-3) = 30`

`=> 12x^2 - 3x - 12x^2 + 6x = 30`

`=> 3x = 30`

`=> x = 30 \div 3`

`=> x=10`

Vậy, `x=10`

`b)`

`2x(3-2x) + 2x(2x-1) = 15`

`=> 6x- 4x^2 + 4x^2 - 2x = 15`

`=> 4x = 15`

`=> x = 15/4`

Vậy, `x=15/4`

`c)`

`(5x-2)(4x-1) + (10x+3)(2x-1) = 1`

`=> 5x(4x-1) - 2(4x-1) + 10x(2x-1) + 3(2x-1)=1`

`=> 20x^2-5x - 8x + 2 + 20x^2 - 10x +6x - 3 =1`

`=> 40x^2 -17x - 1 = 1`

`d)`

`(x+2)(x+2)-(x-3)(x+1)=9`

`=> x^2 + 2x + 2x + 4 - x^2 - x + 3x + 3=9`

`=> 6x + 7 =9`

`=> 6x = 2`

`=> x=2/6 =1/3`

Vậy, `x=1/3`

`e)`

`(4x+1)(6x-3) = 7 + (3x-2)(8x+9)`

`=> 24x^2 - 12x + 6x - 3 = 7 + (3x-2)(8x+9)`

`=> 24x^2 - 12x + 6x - 3 = 7 + 24x^2 +11x - 18`

`=> 24x^2 - 6x - 3 = 24x^2 + 18x -11`

`=> 24x^2 - 6x - 3 - 24x^2 + 18x + 11 = 0`

`=> 12x +8 = 0`

`=> 12x = -8`

`=> x= -8/12 = -2/3`

Vậy, `x=-2/3`

`g)`

`(10x+2)(4x- 1)- (8x -3)(5x+2) =14`

`=> 40x^2 - 10x + 8x - 2 - 40x^2 - 16x + 15x + 6 = 14`

`=> -3x + 4 =14`

`=> -3x = 10`

`=> x= - 10/3`

Vậy, `x=-10/3`

16 tháng 6 2023

Hello các bạn còn đó ko?

a: =1/2x^3*x^2-1/2x^3*6x-1/2x^3*10

=1/2x^5-3x^4-5x^3

b: =-3x^2*5x^3+3x^2*4x^2-3x^2*3x+3x^2*3x

=-15x^5+12x^4-9x^3+9x^2

c: \(=3x\cdot5x^2-3x\cdot2x-3x=15x^3-6x^2-3x\)

d: \(=\dfrac{1}{2}x^2y\cdot2x^3-\dfrac{1}{2}x^2y\cdot\dfrac{2}{5}xy^2-\dfrac{1}{2}x^2y=x^5y-\dfrac{1}{5}x^3y^3-\dfrac{1}{2}x^2y\)

23 tháng 9 2017

Bài 1:

a) \(x^2+5x+6=x^2+2x+3x+6=x\left(x+2\right)+3\left(x+2\right)=\left(x+2\right)\left(x+3\right)\)

b) \(2x^2+5x+3=2x^2+2x+3x+3=2x\left(x+1\right)+3\left(x+1\right)=\left(x+1\right)\left(2x+3\right)\)

c) \(x^2-10x+16=x^2-2x-8x+16=x\left(x-2\right)-8\left(x-2\right)=\left(x-2\right)\left(x-8\right)\)

d) \(4x^2+9x+5=4x^2+4x+5x+5=4x\left(x+1\right)+5\left(x+1\right)=\left(x+1\right)\left(4x+5\right)\)

Bài 2:

không rõ đề --> k lm

26 tháng 9 2017

bai 2 la tim x de cac bieu thuc sau duong

a: \(A\left(x\right)+B\left(x\right)\)

\(=-2x^3+11x^2-5x-\dfrac{1}{5}+2x^3-3x^2-7x+\dfrac{1}{5}\)

\(=8x^2-12x\)

b: C(x)=A(x)-B(x)

\(=-2x^3+11x^2-5x-\dfrac{1}{5}-2x^3+3x^2+7x-\dfrac{1}{5}\)

\(=-4x^3+14x^2+2x-\dfrac{2}{5}\)

4 tháng 11 2017

=>(2x+3).(10x+2)=(5x+2).(4x+5)

=>(2x.10x)+(2x.2)+(3.10x)+(3.2)=(5x.4x)+(5x.5)+(2.4x)+(2.5)

=>20x2+4x+30x+6=20x2+25x+8x+10

=>20x2-20x2+4x-8x+30x-25x=10-6

=>0+4x-8x+30x-25x=4

=>-4x+30x-25x=4

=>26x-25x=4

=>x=4

B)=>(3x-1).(5x-34)=(40-5x).(25-3x)

=>15x2-102x-5x+34=1000-120x-125x+15x2

=>15x2-107x+34=1000-245x+15x2

=>15x2-15x2-107x+245x=1000-34

=>0-107x+245x=966

=>138x=966

=>x=7

A,=>(2x+3).(10x+2)=(5x+2).(4x+5)

=>(2x.10x)+(2x.2)+(3.10x)+(3.2)=(5x.4x)+(5x.5)+(2.4x)+(2.5)

=>20x2+4x+30x+6=20x2+25x+8x+10

=>20x2-20x2+4x-8x+30x-25x=10-6

=>0+4x-8x+30x-25x=4

=>-4x+30x-25x=4

=>26x-25x=4

=>x=4

Bài 2:

a: \(=2x^4-x^3-10x^2-2x^3+x^2+10x=2x^3-3x^3-9x^2+10x\)

b: \(=\left(x^2-15x\right)\left(x^2-7x+3\right)\)

\(=x^4-7x^3+3x^2-15x^3+105x^2-45x\)

\(=x^4-22x^3+108x^2-45x\)

c: \(=12x^5-18x^4+30x^3-24x^2\)

d: \(=-3x^6+2.4x^5-1.2x^4+1.8x^2\)

8 tháng 8 2017

a) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)-33\)

\(20x^2-16x-34=10x^2+3x-34\)

\(10x^2-19x=0\)

\(x\left(10x-19\right)=0\)

\(\Leftrightarrow x=0\)

hoặc \(10x-19=0\)

\(\Leftrightarrow x=\dfrac{19}{10}\)

Vạy ..............

b) \(\dfrac{x-1}{x+5}=\dfrac{6}{7}\)

\(\Leftrightarrow1-\dfrac{x-1}{x+5}=1-\dfrac{6}{7}\)

\(\Leftrightarrow\dfrac{x+5}{x+5}-\dfrac{x-1}{x+5}=\dfrac{7}{7}-\dfrac{6}{7}\)

\(\Leftrightarrow\dfrac{\left(x+5\right)-\left(x-1\right)}{x+5}=\dfrac{1}{7}\)

\(\Leftrightarrow\dfrac{x+5-x+1}{x+5}=\dfrac{1}{7}\)

\(\Leftrightarrow\dfrac{\left(x-x\right)+\left(5+1\right)}{x+5}=\dfrac{1}{7}\)

\(\Leftrightarrow\dfrac{6}{x+5}=\dfrac{1}{7}\)

\(\Leftrightarrow x+5=42\)

\(\Leftrightarrow x=37\)