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28 tháng 10 2020

( 2x + 1 )3 - ( 3x + 2 )2 = ( 2x - 5 )( 4x2 + 10x + 25 ) + 6x( 2x + 1 ) - 9x2

⇔ 8x3 + 12x2 + 6x + 1 - ( 9x2 + 12x + 4 ) = 8x3 - 125 + 12x2 + 6x - 9x2

⇔ 8x3 + 12x2 + 6x + 1 - 9x2 - 12x - 4 = 8x3 + 3x2 + 6x - 125

⇔ 8x3 + 3x2 - 6x - 3 = 8x3 + 3x2 + 6x - 125

⇔ 8x3 + 3x2 - 6x - 3 - 8x3 - 3x2 - 6x + 125 = 0

⇔ -12x + 122 = 0

⇔ -12x = -122

⇔ x = 61/6

31 tháng 10 2021
(3x-2)(2x-4)=1-12x²
23 tháng 8 2018

Bài 1:

\(A=-x^2-2x+9\)

\(A=-\left(x^2+2x-9\right)\)

\(A=-\left(x^2+2x+1-10\right)\)

\(A=-\left(x+1\right)^2+10\)

\(-\left(x+1\right)^2\le0\) với mọi x

\(\Rightarrow-\left(x+1\right)^2+10\le10\)

\(\Rightarrow Amax=10\Leftrightarrow x=-1\)

\(B=-9x^2+6x+25\)

\(B=-\left(9x^2-6x-25\right)\)

\(B=-\left[\left(3x\right)^2-2.3x+1-26\right]\)

\(B=-\left(3x-1\right)^2+26\)

\(-\left(3x-1\right)^2\le0\) với mọi x

\(\Rightarrow-\left(3x-1\right)^2+26\le26\)

\(\Rightarrow Bmax=26\Leftrightarrow3x-1=0\Rightarrow x=\dfrac{1}{3}\)

\(C=-x^2+x+1\)

\(C=-\left(x^2-x-1\right)\)

\(C=-\left(x^2-2x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}-1\right)\)

\(C=-\left(x-\dfrac{1}{2}\right)^2+\dfrac{5}{4}\)

\(-\left(x-\dfrac{1}{2}\right)^2\le0\) với mọi x

\(\Rightarrow-\left(x-\dfrac{1}{2}\right)^2+\dfrac{5}{4}\le\dfrac{5}{4}\)

\(\Rightarrow Cmax=\dfrac{5}{4}\Leftrightarrow x=\dfrac{1}{2}\)

\(D=-2x^2+3x+1\)

\(D=-2\left(x^2-\dfrac{3}{2}x-\dfrac{1}{2}\right)\)

\(D=-2\left(x^2-2.x\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{9}{16}-\dfrac{1}{2}\right)\)

\(D=-2\left(x-\dfrac{3}{4}\right)^2+\dfrac{17}{8}\)

\(-2\left(x-\dfrac{3}{4}\right)^2\le0\) với mọi x

\(\Rightarrow-2\left(x-\dfrac{3}{4}\right)^2+\dfrac{17}{8}\le\dfrac{17}{8}\)

\(\Rightarrow Dmax=\dfrac{17}{8}\Leftrightarrow x=\dfrac{3}{4}\)

\(E=-25x^2-10x+7\)

\(E=-\left(25x^2+10x-7\right)\)

\(E=-\left[\left(5x\right)^2+2.5x+1-8\right]\)

\(E=-\left(5x+1\right)^2+8\)

\(-\left(5x+1\right)^2\le0\) với mọi x

\(\Rightarrow-\left(5x+1\right)^2+8\le8\)

\(\Rightarrow Emax=8\Leftrightarrow5x+1=0\Rightarrow x=-\dfrac{1}{5}\)

Bài 2:

\(A=9x^2+6x+4\)

\(A=\left(3x\right)^2+2.3x+1+3\)

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

\(\left(3x+1\right)^2\ge0\) với mọi x

\(\Rightarrow\left(3x+1\right)^2+3\ge3\)

\(\Rightarrow Amin=3\Leftrightarrow x=-\dfrac{1}{3}\)

\(B=4x^2+4x+12\)

\(B=\left(2x\right)^2+2.2x+1+11\)

\(B=\left(2x+1\right)^2+11\)

\(\left(2x+1\right)^2\ge0\) với mọi x

\(\Rightarrow\left(2x+1\right)^2+11\ge11\)

\(\Rightarrow Bmin=11\Leftrightarrow x=-\dfrac{1}{2}\)

\(C=x^2+x+3\)

\(C=x^2+2x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+3\)

\(C=\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\)

\(\left(x+\dfrac{1}{2}\right)^2\ge0\) với mọi x

\(\Rightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\)

\(\Rightarrow Cmin=\dfrac{11}{4}\Leftrightarrow x=-\dfrac{1}{2}\)

\(D=2x^2+3x+1\)

\(D=2\left(x^2+\dfrac{3}{2}x+\dfrac{1}{2}\right)\)

\(D=2\left(x^2+2.x.\dfrac{3}{4}+\dfrac{9}{16}-\dfrac{9}{16}+\dfrac{1}{2}\right)\)

\(D=2\left(x+\dfrac{3}{4}\right)^2-\dfrac{1}{8}\)

\(2\left(x+\dfrac{3}{4}\right)^2\ge0\) với mọi x

\(\Rightarrow2\left(x+\dfrac{3}{4}\right)^2-\dfrac{1}{8}\ge-\dfrac{1}{8}\)

\(\Rightarrow Dmin=-\dfrac{1}{8}\Leftrightarrow x=-\dfrac{3}{4}\)

\(E=64x^2+16x+3\)

\(E=\left(8x\right)^2+2.8x+1+2\)

\(E=\left(8x+1\right)^2+2\)

\(\left(8x+1\right)^2\ge0\) với mọi x

\(\Rightarrow\left(8x+1\right)^2+2\ge2\)

\(\Rightarrow Emin=2\Leftrightarrow x=-\dfrac{1}{8}\)

\(\left(4-3x\right)\left(10x-5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}4-3x=0\\10x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=4\\10x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{1}{2}\end{cases}}}\)

\(\left(7-2x\right)\left(4+8x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}7-2x=0\\4+8x=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=7\\8x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}}}\)

rồi thực hiện đến hết ... 

Brainchild bé ngây thơ qus e , ko thực hiện đến hết như thế đc đâu :>

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

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

\(2x^2-7x+3-4x^2-4x+3=0\)

\(-2x^2-11x+6=0\)

\(2x^2+11x-6=0\)

\(2x^2+12x-x-6=0\)

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

\(\left(x+6\right)\left(2x-1\right)=0\)

\(x+6=0\Leftrightarrow x=-6\)

\(2x-1=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)

\(3x-2x^2=0\)

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

\(x=0\)

\(2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)

Tự lm tiếp nha 

18 tháng 1 2022

một đòn bẫy dài một mét .đặt ở đâu để có thể dùng 3600n có thể nâng tảng đá nặng 120kg?

a) \(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x-6}{9x^2-4}\)

\(=\dfrac{3x+2-3x+2-3x+6}{\left(3x-2\right)\left(3x+2\right)}\)

\(=\dfrac{-3x+10}{\left(3x-2\right)\left(3x+2\right)}\)

b) \(\dfrac{x+25}{2x^2-50}-\dfrac{x+5}{x^2-5x}-\dfrac{5-x}{2x^2+10x}\)

\(=\dfrac{x+25}{2\left(x-5\right)\left(x+5\right)}-\dfrac{x+5}{x\left(x-5\right)}+\dfrac{x-5}{2x\left(x+5\right)}\)

\(=\dfrac{x^2+25x-2\left(x+5\right)^2+\left(x-5\right)^2}{2x\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{x^2+25x-2x^2-20x-50+x^2-10x+25}{2x\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{-5x-25}{2x\left(x-5\right)\left(x+5\right)}\)

\(=\dfrac{-5\left(x+5\right)}{2x\left(x-5\right)\left(x+5\right)}=\dfrac{-5}{2x\left(x-5\right)}\)

 

c) Ta có: \(\dfrac{1-2x}{2x}-\dfrac{4x}{2x-1}-\dfrac{3}{2x-4x^2}\)

\(=\dfrac{-\left(2x-1\right)^2-8x^2+3}{2x\left(2x-1\right)}\)

\(=\dfrac{-\left(4x^2-4x+1\right)-8x^2+3}{2x\left(2x-1\right)}\)

\(=\dfrac{-4x^2+4x-1-8x^2+3}{2x\left(2x-1\right)}\)

\(=\dfrac{-12x^2+4x+2}{2x\left(2x-1\right)}\)

 

31 tháng 8 2015

Rút gọn hết ta được :

a/ 41x - 17 = -21

=> 41x = -4 => x = 4/41

b/ 34x - 17 = 0 

=> 34x = 17

=> x = 17/34 = 1/2

c/ 19x + 56 = 52 

=> 19x = -4

=> x = -4/19

d/ 20x2 - 16x - 34 = 10x2 + 3x - 34

=> 10x2 - 19x = 0

=> x(10x - 19) = 0

=> x = 0 

hoặc 10x - 19 = 0 => 10x = 19 => x = 19/10

Vậy x = 0 ; x = 19/10

2 tháng 1 2016

Rút gọn hết ta được :

a/ 41x - 17 = -21

=> 41x = -4 => x = 4/41

b/ 34x - 17 = 0

=> 34x = 17

=> x = 17/34 = 1/2

c/ 19x + 56 = 52

=> 19x = -4

=> x = -4/19

d/ 20x 2 - 16x - 34 = 10x 2 + 3x - 34

=> 10x 2 - 19x = 0

=> x(10x - 19) = 0

=> x = 0 hoặc 10x - 19 = 0

=> 10x = 19

=> x = 19/10

Vậy x = 0 ; x = 19/10 

25 tháng 8 2020

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

<=> 3x2 - x - 2 - ( 3x2 + 7x + 2 ) = 7

<=> 3x2 - x - 2 - 3x2 - 7x - 2 = 7

<=> -8x - 4 = 7

<=> -8x = 11

<=> x = -11/8

b) ( 6x + 5 )( 2x + 3 ) - ( 4x + 3 )( 3x - 2 ) = 8

<=> 12x2 + 28x + 15 - ( 12x2 + x - 6 ) = 8

<=> 12x2 + 28x + 15 - 12x2 - x + 6 = 8

<=> 27x + 21 = 8

<=> 27x = -13

<=> x = -13/27 

c) 2x( x + 3 ) - ( x + 1 )( 2x + 1 ) - 5 = 9

<=> 2x2 + 6x - ( 2x2 + 3x + 1 ) - 5 = 9

<=> 2x2 + 6x - 2x2 - 3x - 1 - 5 = 9

<=> 3x - 6 = 9

<=> 3x = 15

<=> x = 5

d) ( 5x + 3 )( 4x - 7 ) - ( 10x + 9 )( 2x - 3 ) = 10

<=> 20x2 - 23x - 21 - ( 20x2 - 12x - 27 ) = 10

<=> 20x2 - 23x - 21 - 20x2 + 12x + 27 = 10

<=> -11x + 6 = 10

<=> -11x = 4

<=> x = -4/11

25 tháng 8 2020

a, \(\left(3x+2\right)\left(x-1\right)-\left(x+2\right)\left(3x+1\right)=7\Leftrightarrow-8x-4=7\Leftrightarrow x=-\frac{11}{8}\)

b, \(\left(6x+5\right)\left(2x+3\right)-\left(4x+3\right)\left(3x-2\right)=8\Leftrightarrow27x+21=8\Leftrightarrow x=-\frac{13}{27}\)

c, \(2x\left(x+3\right)-\left(x+1\right)\left(2x+1\right)-5=9\Leftrightarrow3x-6=9\Leftrightarrow x=5\)

d, \(\left(5x+3\right)\left(4x-7\right)-\left(10x+9\right)\left(2x-3\right)=10\Leftrightarrow-11x+6=10\Leftrightarrow x=-\frac{4}{11}\)

19 tháng 6 2019

\(o,x^2-9x+20=0\)

\(\Leftrightarrow x^2-4x-5x+20=0\)

\(\Leftrightarrow x\left(x-4\right)-5\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x-5=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=5\end{cases}}\)

19 tháng 6 2019

\(n,3x^3-3x^2-6x=0\)

\(\Leftrightarrow3x\left(x^2-x-2\right)=0\)

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

\(\Leftrightarrow3x\left[x\left(x+1\right)-2\left(x+1\right)\right]=0\)

\(\Leftrightarrow3x\left(x+1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\orbr{\begin{cases}3x=0\\x+1=0\end{cases}}\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}\orbr{\begin{cases}x=0\\x=-1\end{cases}}\\x=2\end{cases}}\)