Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) làm gọn hệ số dần: 2x=y
<=>\(\left(2x+3\right)\left(4x^2+3\right)=\) \(\left(y+3\right)\left(y^2+3\right)=-19\)
\(y^3+3y^2+3y+9=-19\Leftrightarrow y^3+3y^2+3y+1=-27\)
\(\Leftrightarrow\left(y+1\right)^3=-3^3\Rightarrow y+1=-3\Rightarrow y=-4\Rightarrow x=-2\)
1.a) 2x4-4x3+2x2
=2x2(x2-2x+1)
=2x2(x-1)2
b) 2x2-2xy+5x-5y
=2x(x-y)+5(x-y)
=(2x+5)(x-y)
2.
a) 4x(x-3)-x+3=0
=>4x(x-3)-(x-3)=0
=>(4x-1)(x-3)=0
=> 2 TH:
*4x-1=0 *x-3=0
=>4x=0+1 =>x=0+3
=>4x=1 =>x=3
=>x=1/4
vậy x=1/4 hoặc x=3
b) (2x-3)^2-(x+1)^2=0
=> (2x-3-x-1).(2x-3+x+1)=0
=>(x-4).(3x-2)=0
=> 2 TH
*x-4=0
=> x=0+4
=> x=4
*3x-2=0
=>3x=0-2
=>3x=-2
=>x=-2/3
vậy x=4 hoặc x=-2/3
a, 2(x+5)=x2+5x
=> 2x+10=x2+5x
=> 0=x2+5x-2x-10
=> x2+3x-10=0
=> x2+5x-2x-10=0
=> x(x+5)-2(x+5)=0
=> (x-2)(x+5)=0
=> x-2 =0 hoặc x+5 =0
=> x=2 hoặc x=-5
b, 4x2-25=(2x-5)(2x+7)
=> (2x)2-52=(2x-5)(2x+7)
=> (2x-5)(2x+5) - (2x-5)(2x+7)=0
=> (2x-5)(2x+5-2x-7)=0
=> (2x-5)(-2)=0
=> 2x-5=0
=> 2x=5
=> x =2,5
c, x3+x=0
=>x(x2+1)=0
=> x=0 hoặc x2+1=0
Mà x2+1 >= 1 nên x=0
d, Hình như là thiếu đề
a,=2x+10=x2+5x
=-x2-2x-5x+10=0
=-x2-7x+10=0
Delta=(-7)2-4.-1.10=89
x1=7+căn89/2 x2=7-căn 89/2
CÁC CÂU KHÁC TỰ GIẢI NHA bạn
1, (x2-x+2)2-(x-2)2=(x2-x+2-x+2)(x2-x+2+x-2)=(x2-2x+4)x2
2,a.x3+4x2-29x+24=0
\(\Leftrightarrow\)x3-3x2+7x2-21x-8x+24=0
\(\Leftrightarrow\)(x3-3x2)+(7x2-21x)-(8x+24)=0
\(\Leftrightarrow\)x2(x-3)+7x(x-3)-8(x-3)=0
\(\Leftrightarrow\)(x-3)(x2-x+8x-8)=0
\(\Leftrightarrow\)(x-3)(x-1)(x+8)=0
\(\Leftrightarrow\)\(\left[\begin{matrix}x-3=0\\x-1=0\\x+8=0\end{matrix}\right.\)\(\left[\begin{matrix}x=3\\x=1\\x=-8\end{matrix}\right.\)
vậy pt có tập nghiệm là S=\(\left\{-8;1;3\right\}\)
b. đặt x2-x=y ta có:
y2-14y+24=0 \(\Leftrightarrow\)(y2-2.7y+49)-25=0 \(\Leftrightarrow\)(y-7)2-52=0 \(\Leftrightarrow\)(y-12)(y-2)=0 \(\Leftrightarrow\left[\begin{matrix}y=12\\y=2\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[\begin{matrix}x^2-x=12\\x^2-x=0\end{matrix}\right.\)\(\Leftrightarrow\left[\begin{matrix}x^2-x-12=0\\x^2-x-2=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}\left(x+3\right)\left(x-4\right)=0\\\left(x-2\right)\left(x+1\right)=0\end{matrix}\right.\)\(\Leftrightarrow\left[\begin{matrix}x=-3\\x=4\\x=2\\x=-1\end{matrix}\right.\)
vậy pt có tập nghiệm là S=\(\left\{-3;-1;2;4\right\}\)
3.ta có : 5x2+5y2+8xy+2x-2y+2=0
\(\Leftrightarrow\)(4x2+8xy+4y2)+(x2+2x+1)+(y2-2y+1)=0
\(\Leftrightarrow\)(2x+2y)2+(x+1)2+(y-1)2=0
lại có (2x+2y)2+(x+1)2+(y-1)2\(\ge\)0 dấu = chỉ sảy ra khi và chỉ khi \(\left\{\begin{matrix}\left(2x+2y\right)^2=0\\\left(x+1\right)^2=0\\\left(y-1\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow\left\{\begin{matrix}2x+2y=0\\x+1=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
vậy x=-1 và y=1
2)
a) \(3x^3-3x=0\)
\(\Leftrightarrow3x\left(x^2-1\right)=0\)
\(\Leftrightarrow3x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x-1=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
Vậy x=0 ; x=-1 ; x=1
b) \(x^2-x+\dfrac{1}{4}=0\)
\(\Leftrightarrow x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0\)
\(\Leftrightarrow x-\dfrac{1}{2}=0\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy \(x=\dfrac{1}{2}\)
1)
a) \(\left(x-2\right)\left(x^2+3x+4\right)\)
\(\Leftrightarrow x^3+3x^2+4x-2x^2-6x-8\)
\(\Leftrightarrow x^3+x^2-2x-8\)
b) \(\left(x-2\right)\left(x-x^2+4\right)\)
\(=x^2-x^3+4x-2x+2x^2-8\)
\(=3x^2-x^3+2x-8\)
c) \(\left(x^2-1\right)\left(x^2+2x\right)\)
\(=x^4+2x^3-x^2-2x\)
d) \(\left(2x-1\right)\left(3x+2\right)\left(3-x\right)\)
\(=\left(6x^2+4x-3x-2\right)\left(3-x\right)\)
\(=18x^2+12x-9x-6-6x^3-4x^2+3x^2+2x\)
\(=17x^2+5x-6-6x^3\)
2a) \(4x^2-1=\left(2x\right)^2-1^2=\left(2x+1\right)\left(2x-1\right)\)
b) \(x^2+16x+64=\left(x+8\right)^2\)
c) \(x^3-8y^3=x^3-\left(2y\right)^3\)
\(=\left(x-2y\right)\left(x^2+2xy+4y^2\right)\)
d) \(9x^2-12xy+4y^2=\left(3x-2y\right)^2\)
a) => 8x^3+12x^2+6x+28=0
=> 8x^3+16x^2-4x^2-8x+14x+28=0
=>8x^2(x+2)-4x(x+2)+14(x+2)=0
=>(x+2)(8x^2-4x+14)=0
=>x=-2 hoặc x=0.25