\(^{x^2}\)- 5x =0
x+x+x +82=-2-x
Những câu hỏi liên quan
3x(x-3)_4x+12=0
x^3-5x=0
(3x-2)^2-(x+2)^2=0
x^2-9-4(x+3)=0
Giải phương trình
6(2x-5)-6x2+15x=0
x+2/x=x^2+5x+4/x^2+2x+x/x+2
a: =>6(2x-5)-3x(2x-5)=0
=>(2x-5)(6-3x)=0
=>x=5/2 hoặc x=2
b: \(\Leftrightarrow\left(x+2\right)^2=x^2+5x+4+x^2\)
\(\Leftrightarrow2x^2+5x+4=x^2+4x+4\)
=>x2+x=0
=>x(x+1)=0
=>x=-1
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Cho A= x^3 -2x+n B=x-2 Tìm n để A chia hết cho B
(x-3)^2-2x+6=0
x^2-5x+6=0
Giúp mình nhanh vs ạ
\(A=x^3-2x+n\)
\(B=n-2\)
\(A\text{⋮}B\) ⇒ \(\left(x^3-2x+n\right)\text{⋮}\left(n-2\right)\)
⇒ \(\left[\left(x^3-2x^2\right)+\left(2x^2-4x\right)+\left(2x-4\right)+\left(n+4\right)\right]\text{⋮}\left(n-2\right)\)
⇒ \(\left[x^2\left(x-2\right)+2x\left(x-2\right)+2\left(x-2\right)+\left(n+4\right)\right]\text{⋮}\left(n-2\right)\)
⇒ \(\left[\left(x-2\right)\left(x^2+2x+2\right)+\left(n+4\right)\right]\text{⋮}\left(x-2\right)\)
Vì \(\left(x-2\right)\left(x^2+2x+2\right)\text{⋮}\left(n-2\right)\)
Để \(A\text{⋮}B\)
⇒ \(n+4=0\)
⇒ \(n=-4\)
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phân tích các đa thức sau thành nhân tử
1) x^2+5x+8
2) x^2+8x+7
3) x^2-6x-16
4) 4x^2-8x+3
5) 3x^2-11x+6
1: Đa thức này ko phân tích được nha bạn
2: \(x^2+8x+7\)
\(=x^2+x+7x+7\)
\(=x\left(x+1\right)+7\left(x+1\right)\)
\(=\left(x+1\right)\left(x+7\right)\)
3: \(x^2-6x-16\)
\(=x^2-8x+2x-16\)
\(=x\left(x-8\right)+2\left(x-8\right)\)
\(=\left(x-8\right)\left(x+2\right)\)
4: \(4x^2-8x+3\)
\(=4x^2-2x-6x+3\)
\(=2x\left(2x-1\right)-3\left(2x-1\right)\)
\(=\left(2x-1\right)\left(2x-3\right)\)
5: \(3x^2-11x+6\)
\(=3x^2-9x-2x+6\)
\(=3x\left(x-3\right)-2\left(x-3\right)\)
\(=\left(x-3\right)\left(3x-2\right)\)
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giải phương trình :
a,x^4+2x^3+5x^2+4x-12=0
b,(x+1)^3+(x-2)^3=(2x-1)^3
c,(x+1)^4+(x+3)^3=82
a) x4 + 2x3 + 5x2 + 4x - 12 = 0
=> x4 - x3+ 3x3- 3x2 + 8x2 -8x + 12x - 12 = 0
=> x3( x - 1) + 3x2( x - 1) + 8x( x - 1) + 12 ( x - 1 ) = 0
=> ( x - 1)( x3 + 3x2 + 8x + 12 ) = 0
=> ( x - 1)( x3 + 2x2 + x2 + 2x + 6x + 12 ) = 0
=> ( x - 1)[ x2( x + 2) + x( x + 2) + 6( x + 2) ] = 0
=> ( x - 1)( x + 2)( x2 + x + 6 ) = 0
Ta thấy : x2 + x + 6
= x2 + 2.\(\dfrac{1}{2}x+\dfrac{1}{4}-\dfrac{1}{4}+6=\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}\text{≥}\dfrac{23}{4}>0\text{∀}x\)
=> ( x - 1)( x + 2 ) = 0
=> x = 1 hoặc x = -2
Vậy,....
b) ( x + 1)3 + ( x - 2)3 = ( 2x - 1)3
=>x3+ 3x2 + 3x + 1 + x3 - 6x2 + 12x - 8 - ( 8x3 - 12x2 + 6x - 1)=0
=> 2x3 - 3x2 + 15x - 7 - 8x3 + 12x2 - 6x + 1 = 0
=> 9x2 - 6x3 + 9x - 6 = 0
=> 9x( x + 1) -6( x3 + 1 ) = 0
=> 9x( x + 1) - 6( x + 1)( x2 - x + 1) = 0
=> 3( x + 1)( 3x - 2x2 + 2x - 2) = 0
=> 3( x + 1)( - 2x2 + 5x - 2) = 0
=> 3( x + 1)( - 2x2 + x + 4x - 2) = 0
=> 3( x + 1)[ x( 1 - 2x ) - 2( 1 - 2x ) ] = 0
=> 3( x + 1)( 1 - 2x )( x - 2) = 0
Suy ra :
* x + 1 = 0 => x = -1
* 1 - 2x = 0 => x = \(\dfrac{1}{2}\)
* x - 2 = 0 => x = 2
Vậy,.....
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Trong các phương trình sau phương trình nào là phương trình bậc nhất một ẩn
A.2/x-3=0 B.0x+2=0
C.5x=x-3 D.0x×3=0
PT bậc nhất 1 ẩn là PT có dạng $ax+b=0$ với $a\neq 0$
Đáp án C. PT có dạng $4x+3=0$
a) 5x-18=72
b) (x-36)-32=0
c) 156 -(x+6)=82
d) 2 mủ x =16
a) \(5x-18=72\)
\(5x=72+18\)
\(5x=100\)
\(x=100\div5\)
\(x=20\)
b) \(\left(x-36\right)-32=0\)
\(x-36=32\)
\(x=32+36\)
\(x=68\)
c) \(156-\left(x+6\right)=82\)
\(x+6=156-82\)
\(x-6=74\)
\(x=74+6\)
\(x=80\)
d) \(2^x=16\)
\(2^x=2^4\)
=> x = 4
30 tháng 7 2021 lúc 20:10
Giair phương trình:1) sqrt[5]{32-x^2}-sqrt[5]{1-x^2}42) sqrt{x}+sqrt[4]{20-x}43) x^3+12sqrt{3x-1}4) sqrt[3]{x-1}+3sqrt[4]{82-x}5) a.left(x+3sqrt{x}+2right)left(x+9sqrt{x}+18right)168xb.sqrt{5x^2+14x+9}-sqrt{x^2-x-20}5sqrt{x+1}
Đọc tiếp
Giair phương trình:
1) \(\sqrt[5]{32-x^2}-\sqrt[5]{1-x^2}=4\)
2) \(\sqrt{x}+\sqrt[4]{20-x}=4\)
3) \(x^3+1=2\sqrt{3x-1}\)
4) \(\sqrt[3]{x-1}+3=\sqrt[4]{82-x}\)
5)
\(a.\left(x+3\sqrt{x}+2\right)\left(x+9\sqrt{x}+18\right)=168x\)
\(b.\sqrt{5x^2+14x+9}-\sqrt{x^2-x-20}=5\sqrt{x+1}\)
a) ĐKXĐ: \(x\ge0\)
Ta có: \(\left(x+3\sqrt{x}+2\right)\left(x+9\sqrt{x}+18\right)=168x\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)\left(\sqrt{x}+6\right)=168x\)
\(\Leftrightarrow\left(x+6\right)^2+12\sqrt{x}\left(x+6\right)-133=0\)
\(\Leftrightarrow\left(x+6\right)^2+19\sqrt{x}\left(x+6\right)-7\sqrt{x}\left(x+6\right)-133=0\)
\(\Leftrightarrow\left(x+6\right)\left(x+19\sqrt{x}+6\right)-7\sqrt{x}\left(x+19\sqrt{x}+6\right)=0\)
\(\Leftrightarrow\left(x-7\sqrt{x}+6\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=36\end{matrix}\right.\)
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ai giúp e với
tìm x :
3x ( x + 1 ) - 2x ( x + 2 ) = - 1 - x
4x ( x - 2019 ) - x + 2019 = 0
( x - 4 )^2 - 36 = 0
x^2 + 8x + 16 = 0
x ( x + 6 ) - 7x - 42 = 0
25x^2 - 9 = 0
\(a,PT\Leftrightarrow3x^2+3x-2x^2-4x=-1-x\Leftrightarrow x^2=-1\left(\text{vô nghiệm}\right)\)
Vậy: ...
\(b,PT\Leftrightarrow4x\left(x-2019\right)-\left(x-2019\right)=0\Leftrightarrow\left(x-2019\right)\left(4x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2019\\x=\dfrac{1}{4}\end{matrix}\right.\)
Vậy: ...
\(c,PT\Leftrightarrow\left(x-4-6\right)\left(x-4+6\right)=0\Leftrightarrow\left(x-10\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)
Vậy: ...
\(d,PT\Leftrightarrow\left(x+4\right)^2=0\Leftrightarrow x=-4\)
Vậy: ...
\(e,PT\Leftrightarrow x\left(x+6\right)-7\left(x+6\right)=0\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)
Vậy: ...
\(f,PT\Leftrightarrow\left(5x-3\right)\left(5x+3\right)=0\Leftrightarrow x=\pm\dfrac{3}{5}\)
Vậy: ...
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