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VT
0
HT
21 tháng 7 2017
1)
a) \(x^2+12x+36=\left(x+6\right)^2\)
b) \(x^2-x+\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2\)
Tick nha
HT
21 tháng 7 2017
3)
a)\(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(\Leftrightarrow x^3+8-x^3-2x=15\)
\(\Leftrightarrow-2x=15-8\)
\(\Leftrightarrow-2x=7\)
\(\Rightarrow x=\dfrac{-7}{2}\)
b) \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2\right)-5x+1=28\)
\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3-10x^2+2x+4x^2-5x+1=28\)
\(\Leftrightarrow0-3x^2+23x+28=28\)
\(\Leftrightarrow-3x^2+23x=0\)
\(\Leftrightarrow-x\left(3x-23\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-x=0\\3x-23=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{23}{3}\end{matrix}\right.\)
c) \(\left(x^2-1\right)^3-\left(x^4+x^2+1\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow x^6-3x^4+3x^2-1-x^6-2x^4-2x^2-1=0\)
\(\Leftrightarrow-5x^4+x^2-2=0\)
Đặt \(-5t^2+t-2=0\)
\(\Delta=1^2-4\left(-5\right)\left(-2\right)=-39< 0\)
\(\Rightarrow PTVN\)
22 tháng 5 2022
a: \(=x^6+27-x^6-9x^4-27x^2-27\)
\(=-9x^4-27x^2\)
b: \(=\left(x^2-1\right)^3-x^6+1\)
\(=x^6-3x^4+3x^2-1-x^6+1\)
\(=-3x^4+3x^2\)
c: \(=5\left(x^2-4\right)-\left(16x^2-24x+9\right)+17\)
\(=5x^2-20-16x^2+24x-9+17\)
\(=-11x^2+24x-12\)
LT
13 tháng 8 2020
a) \(\left(x+2\right)\left(x^2-4x+4\right)-\left(x^3+2x^2\right)=5\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-4x+4\right)-x^2\left(x+2\right)=5\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-4x+4-x^2\right)=5\)
\(\Leftrightarrow\left(x+2\right)\left(4-4x\right)=5\)
\(\Leftrightarrow4x-4x^2+8-8x=5\)
\(\Leftrightarrow-4x^2-4x+3=0\)
\(\Leftrightarrow4x^2+4x-3=0\)
\(\Leftrightarrow4x^2-2x+6x-3=0\)
\(\Leftrightarrow2x\left(2x-1\right)+3\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\2x+3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{3}{2}\end{matrix}\right.\)
Vậy \(x=\left\{\frac{1}{2};-\frac{3}{2}\right\}\)
b) \(6x^2-6x\left(-2+x\right)=36\)
\(\Leftrightarrow6x^2+12x-6x^2=36\)
\(\Leftrightarrow12x=36\)
\(\Leftrightarrow x=3\)
Vậy x = 3
c) \(\left(x+2\right)^2+\left(x-3\right)^2-2\left(x-1\right)\left(x+1\right)=9\)
\(\Leftrightarrow x^2+4x+4+x^2-6x+9-2\left(x^2-1\right)=9\)
\(\Leftrightarrow2x^2-2x+13-2x^2+2=9\)
\(\Leftrightarrow15-2x=9\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=3\)
Vậy x = 3
d) \(\left(x+5\right)^2-9=0\)
\(\Leftrightarrow\left(x+5\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+5\right)^2=3^2\\\left(x+5\right)^2=\left(-3\right)^2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x+5=3\\x+5=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-8\end{matrix}\right.\)
Vậy x ={-2; -8}
e) \(\left(x-2\right)^3=x^3+6x^2=7\) (Câu này sai đề thì phải! Mình sửa lại đề, có gì không giống với đề của bạn thì ib mình sửa nha!)
\(\left(x-2\right)^3-x^3+6x^2=7\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+6x^2=7\)
\(\Leftrightarrow12x-8=7\)
\(\Leftrightarrow12x=15\)
\(\Leftrightarrow x=\frac{5}{4}\)
Vậy \(x=\frac{5}{4}\)
#Chúc bạn học tốt!
14 tháng 9 2020
a) Ta có: \(\left(6x-2\right)^2+\left(5x-2\right)^2-4\left(3x-1\right)\left(5x-2\right)=0\)
\(\Leftrightarrow\left(6x-2\right)^2-2\cdot\left(6x-2\right)\left(5x-2\right)+\left(5x-2\right)^2=0\)
\(\Leftrightarrow\left(6x-2-5x+2\right)^2=0\)
\(\Leftrightarrow x^2=0\)
hay x=0
Vậy: x=0
b) Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=5\)
\(\Leftrightarrow x^3-1-x\left(x^2-4\right)-5=0\)
\(\Leftrightarrow x^3-6-x^2+4x=0\)
\(\Leftrightarrow4x-6=0\)
\(\Leftrightarrow4x=6\)
hay \(x=\frac{3}{2}\)
Vậy: \(x=\frac{3}{2}\)
c) Ta có: \(\left(x-1\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+3\left(x^2-4\right)=2\)
\(\Leftrightarrow x^3-3x^2+3x-1-\left(x^3+27\right)+3x^2-12-2=0\)
\(\Leftrightarrow x^3+3x-15-x^3-27=0\)
\(\Leftrightarrow3x-42=0\)
\(\Leftrightarrow3x=42\)
hay x=14
Vậy: x=14
24 tháng 5 2018
1.\(\left(x-5\right).\left(x+5\right)-\left(x+3\right)^2=2x-3\)
\(\Leftrightarrow x^2-25-\left(x^2+6x+9\right)=2x-3\)
\(\Leftrightarrow x^2-25-x^2-6x-9=2x-3\)
\(\Leftrightarrow x^2-25-x^2-6x-9-2x+3=0\)
\(\Leftrightarrow-8x-31=0\)
\(\Leftrightarrow x=\dfrac{-31}{8}\)
24 tháng 5 2018
\(\left(x-4\right)^3-\left(x-5\right)\left(x^2+5x+25\right)=\left(x+2\right)\left(x^2-2x+4\right)-\left(x+4\right)^3\)
\(\Leftrightarrow\left(x-4\right)^3-\left(x^3-5^3\right)=\left(x^3+2^3\right)-\left(x+4\right)^3\)
\(\Leftrightarrow\left(x-4\right)^3-x^3+5^3=x^3+2^3-\left(x+4\right)^3\)
\(\Leftrightarrow\left(x^3-12x^2+48x-64\right)-x^3+5^3=x^3+2^3-\left(x^3+12x^2+48x+64\right)\)
\(\Leftrightarrow x^3-12x^2+48x-64-x^3+5^3=x^3+2^3-x^3-12x^2-48x-64\)
\(\Leftrightarrow-12x^2+48x-64+5^3=2^3-12x^2-48x-64\)
\(\Leftrightarrow-12x^2+48x-61=-12x^2-48x-56\)
\(\Leftrightarrow96x=-117\)
\(\Leftrightarrow x=\dfrac{-117}{96}=\dfrac{-39}{32}\)
2-x2(x2+x+1)=-x4-x3-x2+m
2-x4-x3-x2=-x4-x3-x2+m (=) m=2
vậy ..
chúc bn hc tốt
\(2-x^2\left(x^2+x+1\right)=-x^4-x^3-x^2+m\)
\(\Leftrightarrow-x^4-x^3-x^2-m=-x^4-x^3-x^2+2\)
\(\Leftrightarrow-x^4-x^3-x^2-m+x^4=-x^4-x^3-x^2+2+x^4\)
\(\Leftrightarrow-x^3-x^2-m=-x^3-x^2+2\)
\(\Leftrightarrow-x^3-x^2-m+x^3=-x^3-x^2+2+x^3\)
\(\Leftrightarrow-x^3-m=-x^2+2\)
\(\Leftrightarrow-x^2-m+x^2=-x^2+2+x^2\)
\(\Leftrightarrow-m=2\)
\(\Leftrightarrow\frac{-m}{-1}=\frac{-2}{-1}\)
\(\Leftrightarrow x=2\)
Vậy: x = 2