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\(3x\left(x-1\right)+2\left(1-x\right)=0.\)
\(\Leftrightarrow3x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-2\right)=0\)
\(\Rightarrow x=1\) hoặc \(x=\frac{2}{3}\)
\(5x\left(x-3\right)=x-3\)
\(\Rightarrow5x\left(x-3\right)-\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(5x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\5x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{1}{5}\end{cases}}}\)
a)\(2x\left(x-2016\right)-2x+4032=0\)
\(\Leftrightarrow2x\left(x-2016\right)-2\left(x-2016\right)=0\)
\(\Leftrightarrow\left(2x-2\right)\left(x-2016\right)=0\)
\(\Leftrightarrow2\left(x-1\right)\left(x-2016\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x-2016=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=2016\end{array}\right.\)
b)\(5x\left(x-3\right)=x-3\)
\(\Leftrightarrow5x\left(x-3\right)-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-3=0\\5x-1=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=3\\x=\frac{1}{5}\end{array}\right.\)
c)\(\left(3x-1\right)^2=\left(x+2\right)^2\)
\(\Leftrightarrow\left(3x-1\right)^2-\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(3x-1+x+2\right)\left[\left(3x-1\right)-\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(4x+1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}4x+1=0\\2x-3=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{4}\\x=\frac{3}{2}\end{array}\right.\)
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
\(x^3+9x^2+27x+26=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2+7x+13\right)=0\Rightarrow x=-2\)
\(x^3+9x^2+27x+26=0\)
\(\Leftrightarrow x^3+9x^2+27x+27=1\)
\(\Leftrightarrow\left(x+3\right)^3=1^3\)
\(\Leftrightarrow x+3=1\Leftrightarrow x=-2\)
(x+5)2 - 7x(x+5)=0
(x+5)2 - 7x - 35 = 0
(x2 + 10x + 25) - 7x - 35 = 0
x2 + 10x + 25 - 7x - 35 = 0
(x2 + 10x - 7x) + 25 - 35 = 0
x(x + 3) + (-10) = 0
suy ra : x(x + 3) = 10
Mà 10 = 2 x 5 = 1 x 10 = (-2) x (-5) = (-1) x (-10)
Nhưng x và x+3 cách nhau 3 đơn vị nên x(x+3) = 2 x 5 = (-2) x (-5)
suy ra x là -5 hoặc 2
Vì |1/4 - x| ≥ 0; |x - y + z| ≥ 0; |2/3 + y| ≥ 0
=> |1/4 - x| + |x - y + z| + |2/3 + y| ≥ 0
Dấu " = " xảy ra <=>. \(\hept{\begin{cases}\frac{1}{4}-x=0\\x-y+z=0\\\frac{2}{3}+y=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{4}\\\frac{1}{4}-y-\frac{2}{3}=0\\y=\frac{-2}{3}\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{4}\\y=\frac{-5}{12}\\z=\frac{-2}{3}\end{cases}}\)
Vậy ....
x2+6x-7x-42=0
x(x+6)-7(x+6)=0
(x+6)(x-7)=0
x=-6 hoac x=7
( nho l ike nha)