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Bài 10:
a) (x+2)2 -x(x+3) + 5x = -20
=> x2 + 4x + 4 - x2 - 3x + 5x = -20
=> 6x = -20 + (-4)
=> 6x = -24
=> x = -4
b) 5x3-10x2+5x=0
=>5x(x2-2x+1)=0
=>5x(x-1)2 =0
=> 5x=0 hoặc (x-1)2=0
=>x=0 hoặc x=1
c) (x2 - 1)3 - (x4 + x2 + 1)(x2 - 1) = 0
=> (x2 - 1)[(x2 - 1)2 - (x4 + x2 + 1)] = 0
<=> (x2 - 1)(x4 - 2x2 + 1 - x4 - x2 - 1) = 0
<=> (x2 - 1)(-3x2) = 0
<=> (x2 - 1)=0 hoặc (-3x2) =0
<=> x2=1 hoặc x2=0
<=> x=−1;1 hoặc x=0
d)
(x+1)3−(x−1)3−6(x−1)2=-19
⇔x3+3x2+3x+1−(x3−3x2+3x−1)−6(x2−2x+1)+19=0
⇔x3+3x2+3x+1−x3+3x2−3x+1−6x2+12x−6+19=0
⇔12x+13=0⇔12x+13=0
⇔12x=-13
⇔x=-23/12
Học tốt nhé:333
Ta có : \(\frac{x-35}{21}+\frac{x-36}{20}>\frac{x-37}{19}+\frac{x-38}{18}\)(1)
\(\Leftrightarrow\left(\frac{x-35}{21}-1\right)+\left(\frac{x-36}{20}-1\right)\)\(-\left(\frac{x-37}{19}-1\right)-\left(\frac{x-38}{18}-1\right)\)\(>0\)
\(\Leftrightarrow\frac{x-56}{21}+\frac{x-56}{20}-\frac{x-56}{19}-\frac{x-56}{18}\)\(>0\)
\(\Leftrightarrow\left(x-56\right)\left(\frac{1}{21}+\frac{1}{20}-\frac{1}{19}-\frac{1}{18}\right)\)\(>0\)
Vì \(\frac{1}{21}+\frac{1}{20}-\frac{1}{19}-\frac{1}{18}< 0\)
\(\Rightarrow x-56< 0\)\(\Leftrightarrow x< 56\)
Vậy tập nghiệm của BPT(1) là \(S=\left\{x\in R|x< 56\right\}\)
\(x^2+3x=0\)
\(\Leftrightarrow x\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
a/ \(x^2-25=0\)
\(\Rightarrow\left(x+5\right)\left(x-5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+5=0\Rightarrow x=-5\\x-5=0\Rightarrow x=5\end{matrix}\right.\)
b/ \(x\left(x+7\right)+x+7=0\)
\(x\left(x+7\right)+\left(x+7\right)=0\)
\(\left(x+7\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+7=0\Rightarrow x=-7\\x+1=0\Rightarrow x=-1\end{matrix}\right.\)
\(\left(x+1\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=2\)
\(\Leftrightarrow x^2+4x+3-x^2-3x+10=2\)
\(\Leftrightarrow x=-11\)
\(a,\Leftrightarrow x^2-4x-x^2+5x=5\Leftrightarrow x=5\\ b,\Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x^2-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=2\end{matrix}\right.\)
a) \(\Rightarrow\left(x-1\right)\left(3x+5\right)=0\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{5}{3}\end{matrix}\right.\)
b) \(\Rightarrow x\left(x-1\right)-2\left(x-1\right)=0\Rightarrow\left(x-1\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
a)x+18=20+58
x+18=78
x=78-18
x=60
b)123-x=36:12
123-x=3
x=123-3
x=120
a) x + 18 = 20 + 58
x + 18 = 78
x = 78 - 18
x = 60