6x^4 - x^3 - 7x^2 + x +1=0 Tìm x
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giúp mik câu này với . tìm x : x^4 + 6x^3 + 7x^2 + 6x + 1 = 0
Tìm x,biết
a) x(x-1) - (x+1)(x+2) = 0
b) (-2x+1)(x-1)+(x-3)(2x+1) = 0
c) -4x^2 -x +5 = 0
d) 5x^3 -2x^2-3x = 0
e) x^3+7x^2+6x = 0
f) x^3 - 5x +4 =0
a) x(x-1) - (x+1)(x+2) = 0
x\(^2\)- x -x\(^{^2}\)-2x +x+2=0
-2x+2=0
-2x=0+2
-2x=2
x=-1
Vậy x bằng -1
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Tìm x,biết:
a) x^2 - 4x -5 = 0
b) 4x^2 + 7x - 11 = 0
c) -7x^2 + 6x + 1 = 0
d) - 10x^2 +7x+3 = 0
a) x2 - 4x - 5 = 0
=> x2 - 5x + x - 5 = 0
=> x(x - 5) + (x - 5) = 0
=> (x + 1)(x - 5) = 0
=> \(\orbr{\begin{cases}x+1=0\\x-5=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-1\\x=5\end{cases}}\)
b) 4x2 + 7x - 11 = 0
=> 4x2 + 11x - 4x - 11 = 0
=> x(4x + 11) - (4x + 11) = 0
=> (x - 1)(4x + 11) = 0
=> \(\orbr{\begin{cases}x-1=0\\4x+11=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=1\\x=-\frac{11}{4}\end{cases}}\)
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c) -7x2 + 6x + 1 = 0
=> -7x2 + 7x - x + 1 = 0
=> -7x(x - 1) - (x - 1) = 0
=> (-7x - 1)(x - 1) = 0
=> \(\orbr{\begin{cases}-7x-1=0\\x-1=0\end{cases}}\)
=> \(\orbr{\begin{cases}-7x=1\\x=1\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{1}{7}\\x=1\end{cases}}\)
d) -10x2 + 7x + 3 = 0
=> -10x2 + 10x - 3x + 3 = 0
=> -10x(x - 1) - 3(x - 1) = 0
=> (-10x - 3)(x - 1) = 0
=> \(\orbr{\begin{cases}-10x-3=0\\x-1=0\end{cases}}\)
=> \(\orbr{\begin{cases}-10x=3\\x=1\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{3}{10}\\x=1\end{cases}}\)
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\(a,x^2-4x-5=0\)
\(\Rightarrow x^2-5x+x-5=0\)
\(\Rightarrow x\left(x-5\right)+\left(x-5\right)=0\)
\(\Rightarrow\left(x-5\right)\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}}\)
\(b,4x^2+7x-11=0\)
\(\Rightarrow4x^2-4x+11x-11=0\)
\(\Rightarrow4x\left(x-1\right)+11\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(4x+11\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\4x+11=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=-\frac{11}{4}\end{cases}}}\)
\(c,-7x^2+6x+1=0\)
\(\Rightarrow-7x^2+7x-x+1=0\)
\(\Rightarrow-7x\left(x-1\right)-\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(-7x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\-7x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{7}\end{cases}}}\)
\(d,-10x^2+7x+3=0\)
\(\Rightarrow-10x^2+10x-3x+3=0\)
\(\Rightarrow-10x\left(x+1\right)-3\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(-10x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+1=0\\-10x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=-\frac{3}{10}\end{cases}}}\)
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1 / (x^2 + 3x + 2 ) + 1 / (x^2 + 5x + 6) + 1 / (x^2 + 7x + 12 ) + . . . + 1/ ( x^2 + 15x + 56 ) = 1/14(x^2 + 6x + 9 ) / (x^2 - 4x + x ) + ( 6x^2 - 36x + 54 ) / (x^2 + 4x + 4) - ( 7x^2 - 63 ) / ( x^2 - 4 ) = 0( 64x^2 + 4 )^2 = ( x^2 - 3x + 5 ) ( 8x - 4 )x^4 - 7x^2 + 4x + 20 =0
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Tìm x biết
a)(x+3)^2(x-2)^2=2x b)7x(x-2)=(x-2) c)8x^3-12x^2+6x-1=0
d)4x^2-9-x(2x-3)=0 e)x^3+5x^2+9x=-45 f)x^3-6x^2-x+30=0
d) \(4x^2-9-x\left(2x-3\right)=0\)
\(\Leftrightarrow4x^2-9-2x^2+3x=0\)
\(\Leftrightarrow2x^2+3x-9=0\)
\(\Delta=3^2-4.2.\left(-9\right)=9+72=81\)
Vậy pt có 2 nghiệm phân biệt
\(x_1=\frac{-3+\sqrt{81}}{4}=\frac{-3}{2}\);\(x_1=\frac{-3-\sqrt{81}}{4}=-3\)
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e) \(x^3+5x^2+9x=-45\)
\(\Leftrightarrow x^3+5x^2+9x+45=0\)
\(\Leftrightarrow x^2\left(x+5\right)+9\left(x+5\right)=0\)
\(\Leftrightarrow\left(x^2+9\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+9=0\\x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\pm3i\\x=-5\end{cases}}\)
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f) \(x^3-6x^2-x+30=0\)
\(\Leftrightarrow\left(x^3-x^2-6x\right)-\left(5x^2-5x-30\right)=0\)
\(\Leftrightarrow x\left(x^2-x-6\right)-5\left(x^2-x-6\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x^2-x-6\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x^2-2x+3x-6\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left[x\left(x-2\right)+3\left(x-2\right)\right]=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow x\in\left\{5;-3;2\right\}\)
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Bài 5: Tìm a , b để các đa thức sau:
1) x^4+6x^3+7x^2-6x+a chia hết cho x2+3x-1
2) x^4-x^3+6x^2-x+a chia hết cho x^2- x+5
3) x^3+3x^2+5x+a chia hết cho x+3
4) x^3+2x^2-7x+a chia hết cho 3x -1
5) 2x^2+ax+1 chia cho x-3 dư 4
3: \(\Leftrightarrow a-15=0\)
hay a=15
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GPT: x^4+6x^3+7x^2-6x+1 = 0
tìm x biết
a,2x^2-6x+4=0
b,5x^2-10x+4=0
c,x^2+7x+12=0
d,13x^2+15x-10=0
g,7x^2-4x-1=0
Tìm x biết
2/7x + 1/2 = -3/4
( 6x + 2/5 ) = -8/125
| x - 2/3 | . ( 18 - 6x mũ 2 ) = 0
Giúp m với m cần gấp ạ please nhanh ạ
Bài giải
a, \(\frac{2}{7}x+\frac{1}{2}=-\frac{3}{4}\)
\(\frac{2}{7}x=-\frac{3}{4}-\frac{1}{2}\)
\(\frac{2}{7}x=-\frac{5}{4}\)
\(x=-\frac{5}{4}\text{ : }\frac{2}{7}\)
\(x=-\frac{35}{8}\)
b, \(\left(6x+\frac{2}{5}\right)=-\frac{8}{125}\)
\(6x=-\frac{8}{125}-\frac{2}{5}\)
\(6x=-\frac{58}{125}\)
\(x=-\frac{58}{125}\text{ : }6\)
\(x=\frac{-29}{375}\)
c, \(\left|x-\frac{2}{3}\right|\cdot\left(18-6x^2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}\left|x-\frac{2}{3}\right|=0\\18-6x^2=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x-\frac{2}{3}=0\\6x^2=18\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x^2=3\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\sqrt{3}\end{cases}}\)
\(\Rightarrow\text{ }x\in\left\{\frac{2}{3}\text{ ; }\sqrt{3}\right\}\)