Tính giá trị biểu thức sau:
a) A=x3+15x2+75x+125 Với x= -10
b) B=4x2+12xy+9y2 Tại x=1;y=2
c) C=27x3-27x2y+9xy2-y3 Với x=1;y= -2
Giải chi tiết giúp mình nha.Cảm ơn
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\(A=x^3+15x^2+75x+125=\left(x+5\right)^3=-125\)
\(B=4x^2+12xy+9y^2=\left(2x+3y\right)^2=\left(3+6\right)^2=81\)
Giải:
\(A=x^3-15x^2+75x-125\)
\(\Leftrightarrow A=x^3-3.x^2.5+3.x.5^2-5^3\)
\(\Leftrightarrow A=\left(x-5\right)^3\)
Tại \(x=35\), giá trị của A là:
\(A=\left(35-5\right)^2=30^2=900\)
Vậy ...
b) Ta có: \(B=x^2+2x+y^2-4y+6\)
\(=x^2+2x+1+y^2-4y+4+1\)
\(=\left(x+1\right)^2+\left(y-2\right)^2+1\ge1\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)
Vậy: \(B_{min}=1\) khi (x,y)=(-1;2)
c) Ta có: \(C=4x^2+4x+9y^2-6y-5\)
\(=4x^2+4x+1+9y^2-6y+1-7\)
\(=\left(2x+1\right)^2+\left(3y-1\right)^2-7\ge-7\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{1}{3}\end{matrix}\right.\)
Vậy: \(C_{min}=-7\) khi \(\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{1}{3}\end{matrix}\right.\)
\(A=2x^2+x=2\left(x^2+\dfrac{1}{2}x\right)=2\left(x^2+2.\dfrac{1}{4}x+\dfrac{1}{16}-\dfrac{1}{16}\right)\)
\(=2\left[\left(x+\dfrac{1}{4}\right)^2-\dfrac{1}{16}\right]\ge-\dfrac{1}{8}\) dấu"=' xảy ra<=>x=\(-\dfrac{1}{4}\)
\(B=x^2+2x+y^2-4y+6\)
\(=x^2+2x+1+y^2-4y+4+1=\left(x+1\right)^2+\left(y-2\right)^2+1\)
\(\ge1\) dấu"=" xảy ra<=>x=-1;y=2
\(C=4x^2+4x+9y^2-6y-5\)
\(=4x^2+4x+1+9y^2-6y+1-7\)
\(=\left(2x+1\right)^2+\left(3y-1\right)^2-7\ge-7\)
dấu"=" xảy ra<=>x=\(-\dfrac{1}{2},y=\dfrac{1}{3}\)
\(D=\left(2+x\right)\left(x+4\right)-\left(x-1\right)\left(x+3\right)^2\)
=\(x^2+6x+8-\left(x-1\right)\left(x+3\right)^2\)
\(=\left(x+3\right)^2-1-\left(x-1\right)\left(x+3\right)^2\)
\(=\left(x+3\right)^2\left(2-x\right)-1\ge-1\)
dấu"=" xảy ra\(< =>\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
\(a,=\left(x+\dfrac{5}{2}\right)^2\\ b,=\left(2x+3y\right)^2\\ c,=a^2+b^2+c^2+2ab-2bc-2ac\\ d,=\left(4x-1\right)^2\\ e,=a^2+b^2+c^2+2ab+2bc+2ac\\ f,=a^2+b^2+c^2-2ab+2bc-2ac\)
a)
\(\left(3x\right)^3-3.\left(3x\right)^2.1+3.3x.2^2-2^3=0\)
\(\left(3x-2\right)^3=0\)
3x-2=0
3x=2
x=2/3
b)
\(x^3-3.x^2.5+3.x.5^2+5^3=0\)
\(\left(x-5\right)^3=0\)
x-5=0
x=5
a. \(x^3+15x^2+75x+125\)\(=x^3+3.x^2.5+3.x.5^2+5^3=\left(x+5\right)^3\)
b. \(x^3-9x^2+27x-27=\)\(x^3-3.x^2.3+3x.3^2-27=\left(x-3\right)^3\)
a)x2-6x+9
=x2-2.x.3+32
=(x-3)2
b)4x2+4x+1
=(2x)2+2.2x.1+12
=(2x+1)2
c)4x2+12xy+9y2
=(2x)2+2.2x.3y+(3y)2
=(2x+3y)2
d)4x4-4x2+4
=(2x2)2-2.2x2.2+22
=(2x2-2)2
\(a,A=\left(x+5\right)^3=\left(-10+5\right)^3=\left(-5\right)^3=-125\\ b,B=\left(2x+3y\right)^2=\left(2\cdot1+3\cdot2\right)^2=7^2=49\\ c,C=\left(3x-y\right)^3=\left(3\cdot1+2\right)^3=5^3=125\)