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\(A=x^3+3x^2+3x+6\)
\(=x^3+3x^2+3x+1+5\)
\(=\left(x+1\right)^3+5\)
Thay x = 19 vào biểu thức \(A=\left(x+1\right)^3+5\)ta được:
\(A=\left(19+1\right)^3+5=20^3+5=8000+5=8005\)
Vậy giá trị của biểu thức A tại x = 19 là 8005.
\(B=x^3-3x^2+3x\)
\(=x^3-3x^2+3x-1+1\)
\(=\left(x-1\right)^3+1\)
Thay x = 11 vào biểu thức \(B=\left(x-1\right)^3+1\)ta được:
\(B=\left(11-1\right)^3+1=10^3+1=1000+1=1001\)
Vậy giá trị của biểu thức B tại x = 11 là 1001.
\(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-2\right)\left(x+1\right)+3x-2=0\)
\(x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+6x+12+3x-2=0\)
\(1+1+6x+3x+12-2=0\)
\(9x+12=0\)
\(9x=-12\)
\(x=\frac{-4}{3}\)
\(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-2\right)\left(x+1\right)+3x-2=0\)
\(\Leftrightarrow\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-2\right)\left(x+1\right)+3x=0+2\)
\(\Leftrightarrow\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-2\right)\left(x+1\right)+3x=2\)
\(\Leftrightarrow9x+14=2\)
\(\Leftrightarrow9x=2-14\)
\(\Leftrightarrow9x=-12\)
\(\Leftrightarrow x=\frac{-12}{9}=\frac{-4}{3}\)
\(\Rightarrow x=\frac{-4}{2}\)
Sửa đề: (sửa sai thì em làm lại:v) \(A=\left(x^2+3x+1\right)^2+\left(3x-1\right)^2-2\left(x^2+3x+1\right)\left(3x-1\right)\)
Đặt \(x^2+3x+1=a;3x-1=b\) cho nó dễ nhìn!
\(A=a^2+b^2-2ab=\left(a-b\right)^2\)
\(=\left(x^2+3x+1-\left(3x-1\right)\right)^2=\left(x^2+2\right)^2=x^4+4x^2+4\)
\(a,=x\left(x^2+9\right)\\ b,=\left(3x+4\right)\left(x^2+9\right)\\ c,=a\left(a+b\right)-M\left(a+b\right)=\left(a-M\right)\left(a+b\right)\\ d,=\left(x+1\right)^2-y^2=\left(x+y+1\right)\left(x-y+1\right)\)
2)
a) \(3x \left(x^2-4\right)=0 \)
\(\Leftrightarrow3x\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x-2=0\\x+2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
Vậy x=0 ; x=2 ; x=-2
b) \(2x^2-x-6=0\)
\(\Leftrightarrow2x^2-4x+3x-6=0\)
\(\Leftrightarrow\left(2x^2-4x\right)+\left(3x-6\right)=0\)
\(\Leftrightarrow2x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x+3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-3}{2}\end{matrix}\right.\)
Vậy x=2 ; \(x=\dfrac{-3}{2}\)
Câu 1 .
a) x3 + x2 + x
= x( x2 + x + 1)
b) xy + y2 - x - y
= x( y - 1) + y( y - 1)
= ( y - 1)( x + y)
2)
a) \(3x\left(x^2-4\right)=0\)
\(\Leftrightarrow3x\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x-2=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
Vậy \(x=0;x=2vàx=-2\)
b) \(2x^2-x-6=0\)
\(\Leftrightarrow2x^2-4x+3x-6=0\)
\(\Leftrightarrow\left(2x^2-4x\right)+\left(3x-6\right)=0\)
\(\Leftrightarrow2x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\2x=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-3}{2}\end{matrix}\right.\)
Vậy \(x=2vàx=\dfrac{-3}{2}\)
1)3.x^2 - 75 = 0
3.x^2 - 3.25 = 0
3.(x^2-25)=0
x^2-5^2=0
(x-5)(x+5)=0
=> x-5=0 hoặc x+5=0
=> x=5 hoặc x=-5
1) \(3x^2-75=0\)
\(\Leftrightarrow3\left(x^2-25\right)=0\)
\(\Leftrightarrow x^2-25=0\)
\(\Leftrightarrow x^2=25\)
\(\Leftrightarrow x=\pm\sqrt{25}=\pm5\)
2) \(x^3+9x^2+27x+27=0\)
\(\Leftrightarrow\left(x+3\right)^3=0\)
\(\Leftrightarrow x+3=0\Leftrightarrow x=-3\)
3) \(x^3+3x^2+3x=0\)
\(\Leftrightarrow x^3+3x^2+3x+1=1\)
\(\Leftrightarrow\left(x+1\right)^3=1^3\)
\(\Leftrightarrow x+1=1\Leftrightarrow x=0\)
Ta có :
\(x^3-3x^2-3x+1=0\)
\(\Leftrightarrow x^3+x^2-4x^2-4x+x+1=0\)
\(\Leftrightarrow x^2\left(x+1\right)-4x\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-4x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x^2-4x+1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-1\\\left(x-2\right)^2-3=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=2\pm\sqrt{3}\end{cases}}\)
Vậy tập nghiệm của phương trình là : \(S=\left\{-1;2+\sqrt{3};2-\sqrt{3}\right\}\)