\(\left(2x-10\right).\left(x+10\right).\left(x+\sqrt{3}\right)=0\)0 (Bài phân tích đa thức thành nhân tử)(ko phải tìm x)
Lam nhanh giup minh nha!
bài 11 phân tích đa thức thành nhân tử
a,\(x^2-xy+x\) b,\(x^2-2xy-4+y^2\) c,\(x^3-x^2-16x+16\)
bài 12 tìm x biết :
a,\(2x\left(x-5\right)-x\left(3+2x\right)=26\) b,\(2\left(x+5\right)-x^2-5x=0\)
bài 11
a) \(x^2-xy+x\\ =x\left(x-y+1\right)\)
b)
\(x^2-2xy-4+y^2\\ =\left(x^2-2xy+y^2\right)-4\\ =\left(x-y\right)^2-4\\ =\left(x-y-2\right)\left(x-y+2\right)\)
c)
\(x^3-x^2-16x+16\\ =x^2\left(x-1\right)-16\left(x-1\right)\\ =\left(x-1\right)\left(x-4\right)\left(x+4\right)\)
bài 12
\(2x\left(x-5\right)-x\left(3+2x\right)=26\)
\(2x^2-10x-3x-2x^2=26\)
\(-13x=26\\ x=-2\)
b)
\(2\left(x+5\right)-x^2-5x=0\\ 2\left(x+5\right)-x\left(x+5\right)=0\\ \left(x+5\right)\left(2-x\right)=0\\ \left[{}\begin{matrix}x+5=0\\2-x=0\end{matrix}\right.\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
bài 1: phân tích đa thức thành nhân tử
a,2x+10y
b,x\(^2+4x+4\)
c,\(x^2-y^2+10y-25\)
bài 2 tìm x, biết
a,\(x^2-3x+x-3=0\)
b,\(2x\left(x-3\right)-\dfrac{1}{2}\left(4x^2-3\right)=0\)
c,\(x^2-\left(x-3\right)\left(2x-5\right)=9\)
\(B1\\ a,2x+10y=2\left(x+5y\right)\\ b,x^2+4x+4=x^2+2.2x+2^2=\left(x+2\right)^2\\ c,x^2-y^2+10y-25\\ =\left(x^2-y^2\right)+5\left(2y-5\right)\\ =\left(x-y\right)\left(x+y\right)+5\left(2y-5\right)\\ B2\)
\(a,x^2-3x+x-3=0\\ =>x\left(x-3\right)+\left(x-3\right)=0\\ =>\left(x+1\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x+1=0\\x-3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\\ b,2x\left(x-3\right)-\dfrac{1}{2}\left(4x^2-3\right)=0\\ =>2x^2-6x-2x^2+\dfrac{3}{2}=0\\ =>-6x=-\dfrac{3}{2}\\ =>x=\left(-\dfrac{3}{2}\right):\left(-6\right)\\ =>x=\dfrac{1}{4}\\ c,x^2-\left(x-3\right)\left(2x-5\right)=9\\ =>x^2-2x^2+6x+5x-15=9\\ =>-x^2+11-15-9=0\\ =>-x^2+11x-24=0\\ =>-x^2+8x+3x-24=0\\ =>-x\left(x-8\right)+3\left(x-8\right)=0\\ =>\left(3-x\right)\left(x-8\right)=0\\ =>\left[{}\begin{matrix}3-x=0\\x-8=0\end{matrix}\right.=>\left[{}\begin{matrix}x=3\\x=8\end{matrix}\right.\)
Phân tích đa thức thành nhân tử :
\(2x^2-3x\sqrt{x+3}+\left(x+3\right)=0\)
\(2x^2-3x\sqrt{x+3}+\left(x+3\right)\)
\(=2x^2-2x\sqrt{x+3}-x\sqrt{x+3}+\left(\sqrt{x+3}\right)^2\)
\(=2x\left(x-\sqrt{x+3}\right)-\sqrt{x+3}\left(x-\sqrt{x+3}\right)\)
\(=\left(2x-\sqrt{x+3}\right)\left(x-\sqrt{x+3}\right)\)
\(2x^2-3x\sqrt{x+3}+\left(x+3\right)\)
\(=2x^2-x\sqrt{x+3}-2x\sqrt{x+3}+\left(\sqrt{x+3}\right)^2\)
\(=x\left(2x-\sqrt{x+3}\right)-\sqrt{x+3}\left(2x-\sqrt{x+3}\right)\)
\(=\left(x-\sqrt{x+3}\right)\left(2x-\sqrt{x+3}\right)\)
bài 1: Phân tích đa thức thành nhân tử
a)\(3x^3+6x^2\)
b)\(x^2-y^2-2x+2y\)
bài 2:
a) tìm x:\(\left(2x-1\right)^2-25=0\)
b) Tìm đa thức Q biết: \(Q.\left(x^2+3x+1\right)=x^3+2x^2-2x-1\)
Gisup mik vs
Cảm ơn
Bài `1:`
`a)3x^3+6x^2=3x^2(x+2)`
`b)x^2-y^2-2x+2y=(x-y)(x+y)-2(x-y)=(x-y)(x+y-2)`
Bài `2:`
`a)(2x-1)^2-25=0`
`<=>(2x-1-5)(2x-1+5)=0`
`<=>(2x-6)(2x+4)=0`
`<=>[(x=3),(x=-2):}`
`b)Q.(x^2+3x+1)=x^3+2x^2-2x-1`
`<=>Q=[x^3+2x^2-2x-1]/[x^2+3x+1]`
`<=>Q=[x^3-x^2+3x^2-3x+x-1]/[x^2+3x+1]`
`<=>Q=[(x-1)(x^2+3x+1)]/[x^2+3x+1]=x-1`
Phân tích đa thức thành nhân tử:
\(x^2+x-30\)
Tìm x:
a) \(\left(x-2\right)^2-x\left(x-5\right)=13\)
b) \(4x^3-100x=0\)
Phân tích đa thức thành nhân tử: \(\left(x+5\right)^2+4\left(x+5\right)\left(x-5\right)+4\left(x^2-10x+25\right)=0\)
\((x+5)^2+4(x+5)(x-5)+4(x^2-10x+25)=0\\\Rightarrow(x+5)^2+4(x+5)(x-5)+4(x^2-2\cdot x\cdot5+5^2)=0\\\Rightarrow(x+5)^2+2\cdot(x+5)\cdot2(x-5)+4(x-5)^2=0\\\Rightarrow(x+5)^2+2\cdot(x+5)\cdot2(x-5)+[2(x-5)]^2=0\\\Rightarrow[(x+5)+2(x-5)]^2=0\\\Rightarrow(x+5+2x-10)^2=0\\\Rightarrow(3x-5)^2=0\\\Rightarrow3x-5=0\\\Rightarrow3x=5\\\Rightarrow x=\frac53\\\text{#}Toru\)
1.Rút gọn rồi tính giá trị của biểu thức sau thành nhân tử
B= \(\left(x-2\right)\left(x+2\right)\left(x+3\right)-\left(x+1\right)^3\)
2.Phân tích đa thức sau thành nhân tử
\(64-x^2-y^2+2xy\)
3. Tìm x
\(2x^3-x^2+2x-1=0\)
1. \(B=\left(x-2\right)\left(x+2\right)\left(x+3\right)-\left(x+1\right)^3\)
\(=\left(x^2-4\right)\left(x+3\right)-\left(x^3+3x^2+3x+1\right)\)
\(=x^3+3x^2-4x-12-x^3-3x^2-3x-1\)
\(=-7x-13\)
2. \(64-x^2-y^2+2xy=64-\left(x^2+y^2-2xy\right)\)
\(=64-\left(x-y\right)^2=\left(8+x-y\right)\left(8-x+y\right)\)
3. \(2x^3-x^2+2x-1=0\)
\(\Leftrightarrow x^2.\left(2x-1\right)+\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x^2+1\right)=0\)
Vì \(x^2\ge0\)\(\Rightarrow x^2+1>0\)
\(\Rightarrow2x-1=0\)\(\Rightarrow2x=1\)\(\Rightarrow x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\)
Bài 1.
B = ( x - 2 )( x + 2 )( x + 3 ) - ( x + 1 )3
= ( x2 - 4 )( x + 3 ) - ( x3 + 3x2 + 3x + 1 )
= x3 + 3x2 - 4x - 12 - x3 - 3x2 - 3x - 1
= -7x - 13
Bài 2.
64 - x2 - y2 + 2xy
= 64 - ( x2 - 2xy + y2 )
= 82 - ( x - y )2
= ( 8 - x + y )( 8 + x - y )
Bài 3.
2x3 - x2 + 2x - 1 = 0
<=> ( 2x3 - x2 ) + ( 2x - 1 ) = 0
<=> x2( 2x - 1 ) + 1( 2x - 1 ) = 0
<=> ( 2x - 1 )( x2 + 1 ) = 0
<=> \(\orbr{\begin{cases}2x-1=0\\x^2+1=0\end{cases}}\Leftrightarrow x=\frac{1}{2}\)( vì x2 + 1 ≥ 1 > 0 ∀ x )
1) \(B=\left(x-2\right)\left(x+2\right)\left(x+3\right)-\left(x+1\right)^3\)
\(B=\left(x^2-4\right)\left(x+3\right)-\left(x^3+3x^2+3x+1\right)\)
\(B=x^2+3x^2-4x-12-x^3-3x^2-3x-1\)
\(B=-7x-13\)
2) \(64-x^2-y^2+2xy\)
\(=64-\left(x^2-2xy+y^2\right)\)
\(=64-\left(x-y\right)^2\)
\(=\left(8-x+y\right)\left(8+x-y\right)\)
3) \(2x^3-x^2+2x-1=0\)
\(\Leftrightarrow x^2\left(2x-1\right)+2x-1=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow2x-1=0\Leftrightarrow x=\frac{1}{2}\)
Phân tích đa thức thành nhân tử
\(5x\left(2x+3\right)+6x+9\)
\(3x\left(x+4\right)+48\left(x+4\right)+5\left(x+4\right)\)
\(5x(2x+3)+6x+9\\=5x(2x+3)+3(2x+3)\\=(2x+3)(5x+3)\)
a: \(5x\left(2x+3\right)+6x+9\)
\(=5x\left(2x+3\right)+\left(6x+9\right)\)
\(=5x\left(2x+3\right)+3\left(2x+3\right)\)
\(=\left(2x+3\right)\left(5x+3\right)\)
b: \(3x\left(x+4\right)+48\left(x+4\right)+5\left(x+4\right)\)
\(=\left(x+4\right)\left(3x+48+5\right)\)
=(x+4)(3x+53)
Bằng cách phân tích vế trái thành nhân tử, giải các PT sau:
a) \(2x.\left(x-3\right)+5\left(x-3\right)\)
b) \(\left(x^2-4\right)+\left(x-2\right).\left(3-2x\right)=0\)
c) \(x^3-3x^2+3x-1=0\)
a: =(x-3)(2x+5)
b: \(\Leftrightarrow\left(x-2\right)\left(x+2+3-2x\right)=0\)
=>(x-2)(5-x)=0
=>x=2 hoặc x=5
c: =>x-1=0
hay x=1