Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Tìm x:
1. \(25x^2-20x+4=0\)
⇔ \(\left(5x-2\right)^2=0\)
⇔ \(5x-2=0\)
⇔ \(5x=2\)
⇔ \(x=\dfrac{2}{5}\)
⇒ S = \(\left\{\dfrac{2}{5}\right\}\)
2. \(\left(2x-3\right)^2-\left(2x+1\right).\left(2x-1\right)=0\)
⇔ \(4x^2-12x+9-\left(4x^2-1\right)=0\)
⇔ \(4x^2-12x+9-4x^2+1=0\)
⇔ \(-12x+10=0\)
⇔ \(-12x=-10\)
⇔ \(x=\dfrac{5}{6}\)
⇒ S \(=\left\{\dfrac{5}{6}\right\}\)
3. \(\left(\dfrac{1}{2}x-1\right)\left(\dfrac{1}{2}x+1\right)-\left(\dfrac{1}{2}x-1\right)^2=0\)
⇔ \(\dfrac{1}{4}x^2-1-\left(\dfrac{1}{4}x^2-x+1\right)=0\)
⇔ \(\dfrac{1}{4}x^2-1-\dfrac{1}{4}x^2+x-1=0\)
⇔ \(-2+x=0\)
⇔ \(x=2\)
⇒ S \(=\left\{2\right\}\)
4. \(\left(2x-3\right)^2+\left(2x+5\right)^2=8\left(x+1\right)^2\)
⇔ \(4x^2-12x+9+4x^2+20x+25=8\left(x^2+2x+1\right)\)
⇔ \(8x^2+8x+34=8x^2+16x+8\)
⇔ \(8x+34=16x+8\)
⇔ \(8x-16x=8-34\)
⇔ \(-8x=-26\)
⇔ \(x=\dfrac{13}{4}\)
⇒ S \(=\left\{\dfrac{13}{4}\right\}\)
5.\(4x^2+12x-7=0\)
⇔ \(4x^2+14x-2x-7=0\)
⇔ \(2x\left(2x+7\right)-\left(2x+7\right)=0\)
⇔ \(\left(2x+7\right)\left(2x-1\right)=0\)
⇔ \(\left[{}\begin{matrix}2x+7=0\\2x-1=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x=\dfrac{-7}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
⇒ S \(=\left\{\dfrac{-7}{2};\dfrac{1}{2}\right\}\)
6. \(\dfrac{1}{4}x^2+\dfrac{2}{3}x-\dfrac{5}{9}=0\)
⇔ \(9x^2+24x-20=0\)
⇔ \(9x^2+30x-6x-20=0\)
⇔ \(3x\left(3x+10\right)-2\left(3x+10\right)=0\)
⇔ \(\left(3x+10\right)\left(3x-2\right)=0\)
⇔ \(\left[{}\begin{matrix}3x+10=0\\3x-2=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x=\dfrac{-10}{3}\\x=\dfrac{2}{3}\end{matrix}\right.\)
⇒ S \(=\left\{\dfrac{-10}{3};\dfrac{2}{3}\right\}\)
7. \(24\dfrac{8}{9}-\dfrac{1}{4}x^2-\dfrac{1}{3}x=0\)
⇔ \(\dfrac{224}{9}-\dfrac{1}{4}x^2-\dfrac{1}{3}x=0\)
⇔ \(896-9x^2-12x=0\)
⇔ \(-896+9x^2+12x=0\)
⇔ \(9x^2+12x-896=0\)
⇔ \(9x^2-84x+96x-896=0\)
⇔ \(3x\left(3x-28\right)+32\left(3x-28\right)=0\)
⇔ \(\left(3x-28\right)\left(3x+32\right)=0\)
⇔ \(\left[{}\begin{matrix}3x-28=0\\3x+32=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x=\dfrac{28}{3}\\x=\dfrac{-32}{3}\end{matrix}\right.\)
⇒ S \(=\left\{\dfrac{-32}{3};\dfrac{28}{3}\right\}\)
a) \(\left(2x-5\right)^2-\left(2x+3\right)\left(2x-3\right)=10\Leftrightarrow\left(4x^2-20x+25\right)-\left(4x^2-9\right)-10=0\)
\(\Leftrightarrow-20x+24=0\Leftrightarrow x=\frac{6}{5}\)
b) \(\left(4x-1\right)\left(x+2\right)-\left(2x+3\right)^2-5\left(x-1\right)=9\Leftrightarrow-10x-15=0\)
\(\Leftrightarrow x=\frac{-3}{2}\)
c) \(\left(x+1\right)^3-\left(x-1\right)^3-2=6\Leftrightarrow\left(x^3+3x^2+3x+1\right)-\left(x^3-3x^2+3x-1\right)-8=0\)
\(\Leftrightarrow6x^2-6=0\Leftrightarrow x=\pm1\)
d) \(\left(x+2\right)\left(x^2-2x+4\right)-\left(x+1\right)\left(x^2-x+1\right)-3\left(-x-2\right)=5\)
\(\Leftrightarrow\left(x^3+8\right)-\left(x^3+1\right)+3x+6=5\Leftrightarrow3x+8=0\Leftrightarrow x=\frac{-8}{3}\)
Bài 1 :
a) \(3x\left(5x^2-2x-1\right)=3x\cdot5x^2+3x\left(-2x\right)+3x\left(-1\right)\)
\(=15x^3-6x^2-3x\)
b) \(\left(x^2-2xy+3\right)\left(-xy\right)\)
\(=x^2\left(-xy\right)-2xy\left(-xy\right)+3\left(-xy\right)\)
\(=-x^3y+2x^2y^2-3xy\)
c) \(\frac{1}{2}x^2y\left(2x^3-\frac{2}{5}xy-1\right)\)
\(=\frac{1}{2}x^2y\cdot2x^3+\frac{1}{2}x^2y\cdot\left(-\frac{2}{5}xy\right)+\frac{1}{2}x^2y\left(-1\right)\)
\(=x^5y-\frac{1}{5}x^3y^2-\frac{1}{2}x^2y\)
d) \(\frac{1}{2}xy\left(\frac{2}{3}x^2-\frac{3}{4}xy+\frac{4}{5}y^2\right)\)
\(=\frac{1}{2}xy\cdot\frac{2}{3}x^2+\frac{1}{2}xy\cdot\left(-\frac{3}{4}xy\right)+\frac{1}{2}xy\cdot\frac{4}{5}y^2\)
\(=\frac{1}{3}x^3y-\frac{3}{8}x^2y^2+\frac{2}{5}xy^3\)
e) \(\left(x^2y-xy+xy^2+y^3\right)\left(3xy^3\right)\)
= \(x^2y\cdot3xy^3-xy\cdot3xy^3+xy^2\cdot3xy^3+y^3\cdot3xy^3\)
\(=3x^3y^4-3x^2y^4+3x^2y^5+3xy^6\)
Bài 2 :
3(2x - 1) + 3(5 - x) = 6x - 3 + 15 - x = (6x - x) - 3 + 15 = 5x - 3 + 15
Thay x = -3/2 vào biểu thức trên ta có : \(5\cdot\left(-\frac{3}{2}\right)-3+15\)
\(=-\frac{15}{2}-3+15=\frac{9}{2}\)
b) 25x - 4(3x - 1) + 7(5 - 2x)
= 25x - 12x + 4 + 35 - 14x
= (25x - 12x - 14x) + 4 + 35 = -x + 4 + 35 = -x + 39
Thay \(x=2\)vào biểu thức trên ta có : -2 + 39 = 37
c) 4x - 2(10x + 1) + 8(x - 2)
= 4x - 20x - 2 + 8x - 16
= (4x - 20x + 8x) - 2 - 16 = -8x - 2 - 16 = -8x - 18
Thay x = 1/2 vào biểu thức trên ta có \(-8\cdot\frac{1}{2}-18=-4-18=-22\)
d) Tương tự
Bài 3:
a) \(2x\left(x-4\right)-x\left(2x+3\right)=4\)
=> 2x2 - 8x - 2x2 - 3x = 4
=> (2x2 - 2x2) + (-8x - 3x) = 4
=> -11x = 4
=> x = \(-\frac{4}{11}\)
b) x(5 - 2x) + 2x(x - 7) = 18
=> 5x - 2x2 + 2x2 - 14x = 18
=> 5x - 14x = 18
=> -9x = 18
=> x = -2
Còn 2 câu làm tương tự
a) \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\)
\(\Leftrightarrow\left(x^2+6x+9\right)-\left(x^2+4x-32\right)-1=0\)
\(\Leftrightarrow2x=-40\)
\(\Rightarrow x=-20\)
b) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=15\)
\(\Leftrightarrow x^3+27-x^3+4x=15\)
\(\Leftrightarrow4x=-12\)
\(\Rightarrow x=-3\)
c) \(\left(x-2\right)^2-\left(x+3\right)^2-4\left(x+1\right)=5\)
\(\Leftrightarrow\left(x^2-4x+4\right)-\left(x^2+6x+9\right)-\left(4x+4\right)=5\)
\(\Leftrightarrow-14x=14\)
\(\Rightarrow x=-1\)
d) \(\left(2x-3\right)\left(2x+3\right)-\left(x-1\right)^2-3x\left(x-5\right)=-44\)
\(\Leftrightarrow4x^2-9-\left(x^2-2x+1\right)-\left(3x^2-15x\right)=-44\)
\(\Leftrightarrow17x=-34\)
\(\Rightarrow x=-2\)
e) \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\)
\(\Leftrightarrow24x=24\)
\(\Rightarrow x=1\)
\(\text{a)}\Rightarrow x-1-x-1-x+2=5\)
\(\Rightarrow-x=5\)
\(\Rightarrow x=-5\)
\(\text{Vậy x=-5}\)
\(\text{b)}\left(2x-1\right)^2-\left(2x+3\right)^2=7\)
\(\Rightarrow\left(4x^2-4x+1\right)-\left(4x^2+12x+9\right)=7\)
\(\Rightarrow4x^2-4x+1-4x^2-12x-9=7\)
\(\Rightarrow-16x-8=7\)
\(\Rightarrow-16x=15\)
\(\Rightarrow x=\frac{-15}{16}\)
\(\text{Vậy }x=\frac{-15}{16}\)
\(\text{c)}\Rightarrow16x^2-9-\left(16x^2-8x+1\right)=8\)
\(\Rightarrow-9+8x-1=8\)
\(\Rightarrow8x=18\)
\(\Rightarrow x=\frac{18}{8}=\frac{9}{4}\)
\(\text{Vậy }x=\frac{9}{4}\)
\(\text{Phần d số rất lẻ, có thể bạn chép sai đề nên mình ko chữa nha~}\)
b, x = -5/3 hoặc x = 4/3.
c, x = 0 hoặc x = 3, -3.
d, x = 0 hoặc x = 2, -2.
e, x = 1 hoặc x = \(\dfrac{-1}{2}\).
a: \(\Leftrightarrow x^2-40x+400-x^2-4x-3=-7\)
=>-44x+397=-7
=>-44x=-404
hay x=101
b: \(\Leftrightarrow\left[{}\begin{matrix}3x+5=0\\4-3x=0\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{5}{3};\dfrac{4}{3}\right\}\)
c: \(\Leftrightarrow x\left(x^2-9\right)=0\)
=>x(x-3)(x+3)=0
hay \(x\in\left\{0;3;-3\right\}\)
d: \(\Leftrightarrow x\left(x-2\right)\left(x+2\right)=0\)
hay \(x\in\left\{0;2;-2\right\}\)
e: =>(2x+1)(1-x)=0
=>x=-1/2 hoặc x=1
a) ( 5x - y )( 25x2 + 5xy + y2 ) = ( 5x )3 - y3 = 125x3 - y3
b) ( x - 3 )( x2 + 3x + 9 ) - ( 54 + x3 ) = x3 - 33 - 54 - x3 = -27 - 54 = -81
c) ( 2x + y )( 4x2 - 2xy + y2 ) - ( 2x - y )( 4x2 + 2xy + y2 ) = ( 2x )3 + y3 - [ ( 2x )3 - y3 ]= 8x3 + y3 - 8x3 + y3 = 2y3
d) ( x + y )2 + ( x - y )2 + ( x + y )( x - y ) - 3x2 = x2 + 2xy + y2 + x2 - 2xy + y2 + x2 - y2 - 3x2 = y2
e) ( x - 3 )3 - ( x - 3 )( x2 + 3x + 9 ) + 6( x + 1 )2
= x3 - 9x2 + 27x - 27 - ( x3 - 33 ) + 6( x2 + 2x + 1 )
= x3 - 9x2 + 27x - 27 - x3 + 27 + 6x2 + 12x + 6
= -3x2 + 39x + 6
= -3( x2 - 13x - 2 )
f) ( x + y )( x2 - xy + y2 ) + ( x - y )( x2 + xy + y2 ) - 2x3
= x3 + y3 + x3 - y3 - 2x3
= 0
g) x2 + 2x( y + 1 ) + y2 + 2y + 1
= x2 + 2x( y + 1 ) + ( y2 + 2y + 1 )
= x2 + 2x( y + 1 ) + ( y + 1 )2
= ( x + y + 1 )2
= [ ( x + y ) + 1 ]2
= ( x + y )2 + 2( x + y ) + 1
= x2 + 2xy + y2 + 2x + 2y + 1
a: \(\dfrac{1-2x}{3}=\dfrac{4x-5}{6}\)
=>\(\dfrac{4x-5}{6}=\dfrac{2-4x}{6}\)
=>4x-5=2-4x
=>8x=7
=>\(x=\dfrac{7}{8}\)
b: \(\left(x-1\right)^2-\left(x+2\right)^2=0\)
=>\(x^2-2x+1-x^2-4x-4=0\)
=>-6x-3=0
=>6x+3=0
=>6x=-3
=>\(x=-\dfrac{1}{2}\)
c: \(\dfrac{2x-1}{6}=\dfrac{7-3x}{2}\)
=>\(\dfrac{3\left(7-3x\right)}{6}=\dfrac{2x-1}{6}\)
=>3(7-3x)=2x-1
=>21-9x=2x-1
=>-11x=-22
=>x=2
d: \(\left(x+3\right)\left(2x-3\right)=2x^2-9\)
=>\(2x^2-3x+6x-9=2x^2-9\)
=>3x=0
=>x=0
Đố thịnh tui ko tin ông có thể làm