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14 tháng 11 2019
\(x^4+5x^3+10x^2+5x-21=0\)
\(\Leftrightarrow x^4-x^3+6x^3-6x^2+16x^2-16x+21x-21=0\)
\(\Leftrightarrow x^3\left(x-1\right)+6x^2\left(x-1\right)+16x\left(x-1\right)+21\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+6x^2+16x+21\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+3x^2+3x^2+9x+21\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x+3\right)+3x\left(x+3\right)+9\left(x+3\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+3\right)\left(x^2+3x+9\right)=0\)
<=> x-1=0 <=> x=1
x+3=0 <=> x=-3
\(x^2+3x+9=x^2+2.\frac{3}{2}x+\frac{9}{4}+\frac{27}{4}=\left(x+\frac{3}{2}\right)^2+\frac{27}{4}>0\)
vậy nghiệm của pt là x=1; x=-3
19 tháng 9 2021
\(1,\\ a,=-35x^5y^4z\\ b,=6x^2-30x-6x^2-3x=-33x\\ c,=x^3-9x^2-2x^2+18x-x+9=x^3-11x^2+17x+9\\ 2,\\ A\left(x\right)+B\left(x\right)=10-2x+4x^3-5x^2-10x^3-5x+6x^2-20\\ =-6x^3+x^2-7x-10\\ A\left(x\right)-B\left(x\right)=10-2x+4x^3-5x^2+10x^3+5x-6x^2+20\\ =14x^3-11x^2+3x+30\\ 3,\\ a,M\left(x\right)=5x+20=0\\ \Leftrightarrow x=-4\\ b,N\left(x\right)=100x^2-49=0\\ \Leftrightarrow\left(10x-7\right)\left(10x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{10}\\x=-\dfrac{7}{10}\end{matrix}\right.\\ c,P\left(x\right)=3x-15=0\\ \Leftrightarrow x=5\)
18 tháng 9 2021
Bài 1;
a)\(5x^3yz.\left(-7x^2y^3\right)=-35.x^5y^4z\)
b)\(6x\left(x-5\right)-x\left(6x+3\right)=6x^2-30x-6x^2-3x=-33x\)
c) \(\left(x-9\right)\left(x^2-2x-1\right)=x^3-2x^2-x-9x^2+18x+9=x^3-11x^2+17x+9\)
MR
1
KT
22 tháng 7 2018
\(x^2-5x-4\left(x-5\right)=0\)
\(\Leftrightarrow\)\(x\left(x-5\right)-4\left(x-5\right)=0\)
\(\Leftrightarrow\)\(\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-5=0\\x-4=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=5\\x=4\end{cases}}\)
Vậy....
\(2x\left(x+6\right)=7x+42\)
\(\Leftrightarrow\)\(2x\left(x+6\right)-7x-42=0\)
\(\Leftrightarrow\)\(2x\left(x+6\right)-7\left(x+6\right)=0\)
\(\Leftrightarrow\)\(\left(x+6\right)\left(2x-7\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+6=0\\2x-7=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-6\\x=\frac{7}{2}\end{cases}}\)
Vậy......
\(x^3-5x^2+x-5=0\)
\(\Leftrightarrow\)\(x^2\left(x-5\right)+\left(x-5\right)=0\)
\(\Leftrightarrow\)\(\left(x-5\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\)\(x-5=0\)
\(\Leftrightarrow\)\(x=5\)
\(x^4-2x^3+10x^2-20x=0\)
\(\Leftrightarrow\)\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\Leftrightarrow\)\(x\left(x-2\right)\left(x^2+10\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
Vậy...
DT
2
20 tháng 10 2019
a) x2 - 3x - x(x + 2) = 2
=> x2 - 3x - x2 - 2x = 2
=> -5x = 2
=> x = -2/5
b) 5x3 - 3x2 + 10x - 6 = 0
=>x2(5x - 3) + 2(5x - 3) = 0
=> (x2 + 2)(5x - 3) = 0
=> \(\orbr{\begin{cases}x^2+2=0\\5x-3=0\end{cases}}\)
=> \(\orbr{\begin{cases}x^2=-2\left(ktm\right)\\5x=3\end{cases}}\)
=> x = 3/5
BA
20 tháng 10 2019
\(a,x^2-3x-x\cdot\left(x+2\right)=2\)
\(x^2-3x-x^2-2x=2\)
\(-5x=2\)
\(x=-\frac{2}{5}\)
\(b,5x^3-3x^2+10x-6=0\)
\(5x\cdot\left(x^2+2\right)-3\cdot\left(x^2+2\right)=0\)
\(\left(x^2+2\right)\cdot\left(5x-3\right)=0\)
\(\hept{\begin{cases}x^2+2=0\\5x-3=0\end{cases}\Rightarrow\hept{\begin{cases}x\notin\varnothing\\x=\frac{3}{5}\end{cases}}}\)
Vậy......
`-10x(2-x)-5x(x+2)=5x(x+3)`
`<=> -20x + 10x^2 - 5x^2 - 10x = 5x^2 +15x`
`<=> 5x^2 - 30x = 5x^2 + 15x`
`<=> -30x = 15x`
`<=> -45x = 0`
`<=> x = 0`
Vậy `S = {0}`
\(-10x\left(2-x\right)-5x\left(x+2\right)=5x\left(x+3\right)\)
\(\text{⇔}10x\left(x-2\right)+5x\left(x-2\right)=-5x\left(x-3\right)\)
\(\text{⇔}\left(x-2\right)\left(10x+5x\right)=-5x\left(x-3\right)\)
\(\text{⇔}15x\left(x-2\right)=-5x^2+15\)
\(\text{⇔}15x^2-30=-5x^2+15\)
\(\text{⇔}15x^2+5x^2=30+15\)
\(\text{⇔}20x^2=45\)
\(\text{⇔}x=\sqrt{\dfrac{45}{20}}=\dfrac{3}{2}\)
Vậy: \(x=\dfrac{3}{2}\)