tính
a)x^2-4x+2
b)x^2+x+2
c)x^2-x+3
d)x^2-6x+5
e)x^2-4x+5
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
23 tháng 7 2021
Bài 2:
a) Ta có: \(A=\left(7x+5\right)^2+\left(3x-5\right)^2-\left(10-6x\right)\left(5+7x\right)\)
\(=\left(7x+5\right)^2+2\cdot\left(7x+5\right)\cdot\left(3x-5\right)+\left(3x-5\right)^2\)
\(=\left(7x+5+3x-5\right)^2\)
\(=\left(10x\right)^2=100x^2\)
Thay x=-2 vào A, ta được:
\(A=100\cdot\left(-2\right)^2=100\cdot4=400\)
b) Ta có: \(B=\left(2x+y\right)\left(y^2-2xy+4x^2\right)-8x\left(x-1\right)\left(x+1\right)\)
\(=8x^3+y^3-8x\left(x^2-1\right)\)
\(=8x^3+y^3-8x^3+8x\)
\(=8x+y^3\)
Thay x=-2 và y=3 vào B, ta được:
\(B=-2\cdot8+3^3=-16+27=11\)
2 tháng 1 2023
a.(x+10) /(4*x)-8* 4 -(2*x)/x+2
-(127*x-10)/(4*x)
(5/2-127*x/4)/x
6 tháng 11 2021
\(A=\left(x^2-4x+4\right)+4=\left(x-2\right)^2+4\ge4\)
\(minA=4\Leftrightarrow x=2\)
\(B=\left(4x^2-12x+9\right)+2=\left(2x-3\right)^2+2\ge2\)
\(minB=2\Leftrightarrow x=\dfrac{3}{2}\)
\(C=3\left(x^2+2x+1\right)-8=3\left(x+1\right)^2-8\ge-8\)
\(minC=-8\Leftrightarrow x=-1\)
\(D=-\left(x^2-2x+1\right)-4=-\left(x-1\right)^2-4\le-4\)
\(maxD=-4\Leftrightarrow x=1\)
\(E=-\left(4x^2-6x+\dfrac{9}{4}\right)-\dfrac{11}{4}=-\left(2x-\dfrac{3}{2}\right)^2-\dfrac{11}{4}\le-\dfrac{11}{4}\)
\(maxA=-\dfrac{11}{4}\Leftrightarrow x=\dfrac{3}{4}\)
\(F=-2\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)-\dfrac{55}{8}=-2\left(x-\dfrac{1}{4}\right)^2-\dfrac{55}{8}\le-\dfrac{55}{8}\)
\(maxF=-\dfrac{55}{8}\Leftrightarrow x=\dfrac{1}{4}\)
\(G=\left(x^2-4xy+4y^2\right)+\left(y^2+y+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(x-2y\right)^2+\left(y+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)
\(maxG=\dfrac{3}{4}\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-\dfrac{1}{2}\end{matrix}\right.\)
\(H=-\left(x^2-2x+1\right)-\left(y^2+4y+4\right)+16=-\left(x-1\right)^2-\left(y+2\right)^2+16\le16\)
\(maxH=16\Leftrightarrow\) \(\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
BP
2
7 tháng 10 2021
b: Ta có: \(\left(2x+7\right)^2=9\left(x+2\right)^2\)
\(\Leftrightarrow\left(3x+4-2x-7\right)\left(3x+4+2x+7\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(5x+11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{11}{5}\end{matrix}\right.\)
7 tháng 10 2021
c: ta có: \(\left(x+2\right)^2=9\left(x^2-4x+4\right)\)
\(\Leftrightarrow\left(3x-6\right)^2-\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(3x-6-x-2\right)\left(3x-6+x+2\right)=0\)
\(\Leftrightarrow\left(2x-8\right)\left(4x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\end{matrix}\right.\)
17 tháng 10 2021
a: \(\left(2x-1\right)^2-3\left(x-1\right)\left(x+2\right)-\left(x-3\right)^2\)
\(=4x^2-4x+1-x^2+6x-9-3\left(x^2+x-2\right)\)
\(=3x^2+2x-8-3x^2-3x+6\)
=-x+2
b: \(\left(x-2\right)\left(2x-1\right)-3\left(x+1\right)^2-4x\left(x+2\right)\)
\(=2x^2-x-4x+2-3x^2-6x-3-4x^2-8x\)
\(=-5x^2-19x-1\)
a)x^2-4x+2=(x-4)2-2=(x-4-\(\sqrt{2}\))(x-4+\(\sqrt{2}\))
b)x^2+x+2=(x+\(\frac{1}{4}\))2+\(\frac{3}{4}\)
c)x^2-x+3=(x-\(\frac{1}{4}\))2+\(\frac{13}{4}\)
d)x^2-6x+5=(x-5)(x-1)
e)x^2-4x+5+(x-4)2+1