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\(1,\\ a,=\left[x^3\left(x-2\right)-4x\left(x-2\right)\right]:\left(x^2-4\right)\\ =x\left(x^2-4\right)\left(x-2\right):\left(x^2-4\right)=x\left(x-2\right)\\ b,=\left(2014-14\right)^2=2000^2=4000000\\ 2,\\ A=2015\cdot2013\cdot\left(2014^2+1\right)\\ A=\left(2014^2-1\right)\left(2014^2+1\right)\\ A=2014^4-1< B=2014^4\)
Bài làm:
Ta có: \(4x^2-4x-3=0\)
\(\Leftrightarrow\left(4x^2-4x+1\right)-4=0\)
\(\Leftrightarrow\left(2x-1\right)^2-2^2=0\)
\(\Leftrightarrow\left(2x-3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=0\\2x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=-\frac{1}{2}\end{cases}}\)
Ta có : \(4x^2-4x-3=0\)
\(\Leftrightarrow\left(4x^2-4x+1\right)-4=0\)
\(\Leftrightarrow\left(2x-1\right)^2=4\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=2\\2x-1=-2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{3}{2}\\x=-\frac{1}{2}\end{cases}}\)
Vậy \(x\in\left\{\frac{3}{2};-\frac{1}{2}\right\}\)
1) theo đề bài ta có:\(\left(2^x-8\right)^3+\left(4^x+13\right)^3+\left(-4^x-2^x-5\right)^3=0\)
Đặt 2^x-8=a;4^x+13=b; -4^x-2^x-5=c
=> a+b+c=0=> a^3+b^3+c^3=3abc=0
=> 3(2^x-8)(4^x+13)(-4^x-2^x-5)=0
=> 2^x-8=0;4^x+13=0;-4^x-2^x-5=0
tìm được x=3
2)ta có\(x^2-2xy+2y^2-2x+6y+5=0\)
<=>\(\left(x^2+y^2+1-2xy-2x+2y\right)+\left(y^2+4y+4\right)=0\)
<=>\(\left(x-y-1\right)^2+\left(y+2\right)^2=0\)
<=> (x-y-1)^2=0 và (y+2)^2=0
=> x=-1;y=-2
\(\frac{2x-1}{x}+\frac{3-x}{4}=2\)
\(ĐKXĐ:x\ne0\)
\(MTC:4x\)
\(\frac{4\left(2x-1\right)}{4x}+\frac{x\left(3-x\right)}{4x}=\frac{8x}{4x}\)
\(\Rightarrow4\left(2x-1\right)+x\left(x-3\right)=8x\)
\(\Leftrightarrow8x-4+x^2-3x=8x\)
\(\Leftrightarrow8x-4+x^2-3x-8x=0\)
\(\Leftrightarrow x^2-3x-4=0\)
\(\Leftrightarrow x^2-4x+x-4=0\)
\(\Leftrightarrow\left(x^2-4x\right)+\left(x-4\right)=0\)
\(\Leftrightarrow x\left(x-4\right)+\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+1\right)=0\)
Hoặc\(\hept{\begin{cases}x-4=0\\x+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=4\left(N\right)\\x=-1\left(N\right)\end{cases}}}\)
Vậy tập nghiệp của pt là \(S=\left\{-1;4\right\}\)
1: Ta có: \(\left(3-x\right)^2+\left(2x+1\right)^2-\left(2-x\right)^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left(x-3\right)^2-\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(x-3+x-2\right)=0\)
\(\Leftrightarrow x=\dfrac{5}{2}\)
2: Ta có: \(\left(1-2x\right)^2-3\left(x-1\right)^2+\left(x+1\right)^2-\left(x-1\right)^2-\left(x-1\right)^2=0\)
\(\Leftrightarrow4x^2-4x+1-3x^2+6x-3+\left(x+1\right)^2-2\left(x-1\right)^2=0\)
\(\Leftrightarrow x^2+2x-2+x^2+2x+1-2\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow2x^2+4x+1-2x^2+4x-2=0\)
\(\Leftrightarrow x=\dfrac{1}{8}\)
( a + 2 )3 - a( a - 3 )2
= a3 + 6a2 + 12a + 8 - a( a2 - 6a + 9 )
= a3 + 6a2 + 12a + 8 - a3 + 6a2 - 9a
= 12a2 + 3a + 8
cách của symbolab:
\(\left(a+2\right)^3-a\left(a-3\right)^2\)
\(=a^3+6a^2+12a+8-a\left(a-3\right)^2\)
\(=a^3+6a^2+12a+8-a\left(a^2-6a+9\right)\)
\(=a^3+6a^2+12a+8-a^3+6a^2-9a\)
\(=12a^2+3a+8\)
x2yz3 = 43
xy2 = 49
=> x3y3z3= 412=> (xyz)3= (44)3=> xyz = 44=256