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c: Ta có: \(x^3+3x^2+3x-7=0\)
\(\Leftrightarrow x+1=2\)
hay x=1
b: Ta có: \(x\left(x-3\right)-4x+12=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
a) \(x^3\)\(-\)\(\frac{1}{4}x\)\(=\)\(0\)
\(x\left(x^2-\frac{1}{4}\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x^2-\frac{1}{4}=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x^2=0,5^2\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=+-0,5\end{cases}}\)
Vậy .............................
b) \(\left(2x-1\right)^2\)\(-\)\(\left(x+3\right)^2\)\(=\)\(0\)
\(\left(2x-1+x+3\right)\left(2x-1-x-3\right)=0\)
\(\left(3x+2\right)\left(x-4\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}3x+2=0\\x-4=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}3x=-2\\x=4\end{cases}}\)\(\orbr{\begin{cases}x=\frac{-2}{3}\\x=4\end{cases}}\)
Vậy ................................
c) \(x^2\)\(\left(x-3\right)\)\(+\)\(12\)\(-\)\(4x\)\(=\)\(0\)
\(x^2\)\(\left(x-3\right)\)\(-\)\(4\)\(\left(x-3\right)\)\(=\)\(0\)
\(\left(x^2-4\right)\left(x-3\right)\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x^2\\x-3=0\end{cases}-4=0}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x^2\\x=3\end{cases}=2^2}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=+-2\\x=3\end{cases}}\)
a)\(x^3-\frac{1}{4}x=0\)
\(\Leftrightarrow x\left(x^2-\frac{1}{4}\right)=0\)
\(\Leftrightarrow x\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\x-\frac{1}{2}=0\\x+\frac{1}{2}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}}\)
a: \(x^3-4x^2-x+4=0\)
=>\(\left(x^3-4x^2\right)-\left(x-4\right)=0\)
=>\(x^2\left(x-4\right)-\left(x-4\right)=0\)
=>\(\left(x-4\right)\left(x^2-1\right)=0\)
=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)
b: Sửa đề: \(x^3+3x^2+3x+1=0\)
=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)
=>\(\left(x+1\right)^3=0\)
=>x+1=0
=>x=-1
c: \(x^3+3x^2-4x-12=0\)
=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)
=>\(x^2\cdot\left(x+3\right)-4\left(x+3\right)=0\)
=>\(\left(x+3\right)\left(x^2-4\right)=0\)
=>\(\left(x+3\right)\left(x-2\right)\left(x+2\right)=0\)
=>\(\left[{}\begin{matrix}x+3=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\\x=-2\end{matrix}\right.\)
d: \(\left(x-2\right)^2-4x+8=0\)
=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)
=>\(\left(x-2\right)^2-4\left(x-2\right)=0\)
=>\(\left(x-2\right)\left(x-2-4\right)=0\)
=>(x-2)(x-6)=0
=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)
a) Ta có: \(Q=-x^2-y^2+4x-4y+2=-\left(x^2+y^2-4x+4y-2\right)\)
\(=-\left(x^2-4x+4+y^2+4y+4\right)+10\)
\(=-\left[\left(x-2\right)^2+\left(y+2\right)^2\right]+10\le10\forall x,y\)
Vậy MaxQ=10 khi x=2, y=-2
b) +Ta có: \(A=-x^2-6x+5=-\left(x^2+6x-5\right)=-\left(x^2+6x+9-14\right)\)
\(=-\left(x^2+6x+9\right)+14=-\left(x+3\right)^2+14\le14\forall x\)
Vậy MaxA=14 khi x=-3
+Ta có: \(B=-4x^2-9y^2-4x+6y+3=-\left(4x^2+9y^2+4x-6y-3\right)\)
\(=-\left(4x^2+4x+1+9y^2-6y+1-5\right)\)
\(=-\left[\left(2x+1\right)^2+\left(3y-1\right)^2\right]+5\le5\forall x,y\)
Vậy MaxB=5 khi x=-1/2, y=1/3
c) Ta có: \(P=x^2+y^2-2x+6y+12=x^2-2x+1+y^2+6y+9+2\)
\(=\left(x-1\right)^2+\left(y+3\right)^2+2\ge2\forall x,y\)
Vậy MinP=2 khi x=1, y=-3
Câu 1:
Phần a đề sai nên mk sửa lại:
a, x2 + 5x - 14 = x2 - 2x + 7x - 14 = x(x - 2) + 7(x - 2) = (x - 2)(x + 7)
b, xz + yz - 5(x + y) = z(x + y) - 5(x + y) = (x + y)(z - 5)
Câu 2:
x2 - 4x = -4
\(\Leftrightarrow\) x2 - 4x + 4 = 0
\(\Leftrightarrow\) (x - 2)2 = 0
\(\Leftrightarrow\) x - 2 = 0
\(\Leftrightarrow\) x = 2
Vậy x = 2
Chúc bn học tốt!
\(x^3-3x^2-4x+12=0\)
\(\Rightarrow x^2\left(x-3\right)-4\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(x^2-4\right)=0\)
\(\Rightarrow\left(x-3\right)\left(x-2\right)\left(x+2\right)=0\)
Tìm được x = 3, x = 2 và x = -2
\(x^4+x^3-4x-4=0\)
\(\Rightarrow x^3\left(x+1\right)-4\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x^3-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+1=0\\x^3=4\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=\sqrt[3]{4}\end{cases}}}\)
Chúc bạn học tốt.