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P = (x + 1)(x + 2)(x + 3)(x + 4) - 32
Q = x2 - 2xy + y2 + 3x - 3y + 1
R = 4x2 + \(\dfrac{1}{x^2}\)
#Hỏi cộng đồng OLM
#Toán lớp 8
b: \(=\left(x-y\right)^2+3\left(x-y\right)+1\)
\(=\left(x-y+\dfrac{3}{2}\right)^2-\dfrac{5}{4}>=-\dfrac{5}{4}\)
Dấu '=' xảy ra khi x-y=-3/2
=>x=y-3/2
c: \(4x^2+\dfrac{1}{x^2}\ge2\sqrt{4x^2\cdot\dfrac{1}{x^2}}=4\)
Dấu '=' xảy ra khi \(4x^4=1\)
=>x4=1/4
hay \(x=\pm\dfrac{\sqrt{2}}{2}\)
nhiều quá bạn ạ
hay bạn tìm hiểu cách thức chung làm dạng bài tìm GTNN chứ như thế này thì làm lâu lắm
a: \(=49x^2-28x+4+21=\left(7x-2\right)^2+21>=21\)
Dấu '=' xảy ra khi x=2/7
b: \(=8\left(x^2-\dfrac{7}{2}x-\dfrac{1}{8}\right)\)
\(=8\left(x^2-2\cdot x\cdot\dfrac{7}{4}+\dfrac{49}{16}-\dfrac{51}{16}\right)\)
\(=8\left(x-\dfrac{7}{4}\right)^2-\dfrac{51}{2}>=-\dfrac{51}{2}\)
Dấu '=' xảy ra khi x=7/4
c: \(C=\left(2x^2+5\right)^2+10>=25+10=35\)
Dấu '=' xảy ra khi x=0
Bài 2:
a) \(\dfrac{x}{x-3}+\dfrac{9-6x}{x^2-3x}\)
\(=\dfrac{x}{x-3}+\dfrac{9-6x}{x\left(x-3\right)}\)
\(=\dfrac{x^2-6x+9}{x\left(x-3\right)}\)
\(=\dfrac{\left(x-3\right)^2}{x\left(x-3\right)}\)
\(=\dfrac{x-3}{x}\)
b) \(\dfrac{6x-3}{x}:\dfrac{4x^2-1}{3x^2}\)
\(=\dfrac{6x-3}{x}.\dfrac{3x^2}{4x^2-1}\)
\(=\dfrac{3\left(2x-1\right).3x^2}{x\left(2x-1\right)\left(2x+1\right)}\)
\(=\dfrac{9x}{2x+1}\)
c) \(\dfrac{x+2}{3x}+\dfrac{x-5}{5x}-\dfrac{x+8}{4x}\)
\(=\dfrac{20x\left(x+2\right)+12x\left(x-5\right)-15x\left(x+8\right)}{60x}\)
\(=\dfrac{20x^2+40x+12x^2-60x-15x^2-120x}{60x}\)
\(=\dfrac{17x^2-140x}{60x}\)
d) \(\dfrac{x^2-x+1}{x^2+x}.\dfrac{x+1}{3x-2}.\dfrac{9x-6}{x^2-x+1}\)
\(=\dfrac{x^2-x+1}{x\left(x+1\right)}.\dfrac{x+1}{3x-2}.\dfrac{3\left(3x-2\right)}{x^2-x+1}\)
\(=\dfrac{3\left(x^2-x+1\right)\left(x+1\right)\left(3x-2\right)}{x\left(x+1\right)\left(3x-2\right)\left(x^2-x+1\right)}\)
\(=\dfrac{3}{x}\).
1) \(\left(x-3\right)\left(x-5\right)+44\)
\(=x^2-3x-5x+15+44\)
\(=x^2-8x+59\)
\(=x^2-2.x.4+4^2+43\)
\(=\left(x-4\right)^2+43\ge43>0\)
\(\rightarrowĐPCM.\)
2) \(x^2+y^2-8x+4y+31\)
\(=\left(x^2-8x\right)+\left(y^2+4y\right)+31\)
\(=\left(x^2-2.x.4+4^2\right)-16+\left(y^2+2.y.2+2^2\right)-4+31\)
\(=\left(x-4\right)^2+\left(y+2\right)^2+11\ge11>0\)
\(\rightarrowĐPCM.\)
3)\(16x^2+6x+25\)
\(=16\left(x^2+\dfrac{3}{8}x+\dfrac{25}{16}\right)\)
\(=16\left(x^2+2.x.\dfrac{3}{16}+\dfrac{9}{256}-\dfrac{9}{256}+\dfrac{25}{16}\right)\)
\(=16\left[\left(x+\dfrac{3}{16}\right)^2+\dfrac{391}{256}\right]\)
\(=16\left(x+\dfrac{3}{16}\right)^2+\dfrac{391}{16}>0\)
-> ĐPCM.
4) Tương tự câu 3)
5) \(x^2+\dfrac{2}{3}x+\dfrac{1}{2}\)
\(=x^2+2.x.\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{9}+\dfrac{1}{2}\)
\(=\left(x+\dfrac{1}{3}\right)^2+\dfrac{7}{18}>0\)
-> ĐPCM.
6) Tương tự câu 5)
7) 8) 9) Tương tự câu 3).
a; \(=x^5-2x^4-x^3-x^3-x^2=x^5-2x^4-2x^3-x^2\)
b: \(=2x^3-6x^2+x^2-3x+x-3\)
\(=2x^3-5x^2-2x-3\)
c: \(=6x^3y^2-3x^3+3x^2-2x^2y^3+x^2y-xy\)
d: \(=x^3-x^3y+x^3y-x^2y^2+xy^3-y^4\)
\(=x^3-x^2y^2+xy^3-y^4\)
a: (x-3)(x-2)<0
=>x-2>0 và x-3<0
=>2<x<3
b: \(\left(x+3\right)\left(x+4\right)\left(x^2+2\right)\ge0\)
\(\Leftrightarrow\left(x+3\right)\left(x+4\right)\ge0\)
=>x>=-3 hoặc x<=-4
c: \(\dfrac{x-1}{x-2}\ge0\)
nên \(\left[{}\begin{matrix}x-2>0\\x-1\le0\end{matrix}\right.\Leftrightarrow x\in(-\infty;1]\cup\left(2;+\infty\right)\)
d: \(\dfrac{x+3}{2-x}\ge0\)
\(\Leftrightarrow\dfrac{x+3}{x-2}\le0\)
hay \(x\in[-3;2)\)
a: \(A=4\cdot15^2-70^2=-4000\)
b: \(B=x^2+2x\left(y+1\right)+\left(y+1\right)^2\)
\(=\left(x+y+1\right)^2\)
\(=100^2=10000\)
c: \(C=b^2-3b+a^2+3a-2ab\)
\(=\left(a-b\right)^2+3\left(a-b\right)\)
\(=\left(a-b\right)\left(a-b+3\right)\)
\(=\left(-5\right)\cdot\left(-5+3\right)=\left(-5\right)\cdot\left(-2\right)=10\)
d: \(D=\left(x-y\right)^3+3xy\left(x-y\right)+3xy\)
\(=\left(-1\right)^3-3xy+3xy\)
=-1
\(a.P=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-32\)
\(P=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-32\)
Đặt : \(x^2+5x+5=t\) , ta có :
\(\left(t-1\right)\left(t+1\right)-32=t^2-1-32=t^2-33=\left(t-\sqrt{33}\right)\left(t+\sqrt{33}\right)\)
Thay : \(x^2+5x+5=t\) , ta có :
\(\left(x^2+5x+5-\sqrt{33}\right)\left(x^2+5x+5+\sqrt{33}\right)\)
\(b.Q=x^2-2xy+y^2+3x-3y+1=\left(x-y\right)^2-3\left(x-y\right)+1=\left(x-y\right)^2-2.\dfrac{3}{2}\left(x-y\right)+\dfrac{9}{4}+1-\dfrac{9}{4}=\left(x-y-\dfrac{3}{2}\right)^2-\dfrac{5}{4}=\left(x-y-\dfrac{3}{2}-\dfrac{\sqrt{5}}{2}\right)\left(x-y-\dfrac{3}{2}+\dfrac{\sqrt{5}}{2}\right)=\left(x-y-\dfrac{3+\sqrt{5}}{2}\right)\left(x-y+\dfrac{\sqrt{5}-3}{2}\right)\)
\(c.R=4x^2+\dfrac{1}{x^2}-20=4x^2-2.2x.\dfrac{1}{x}+\dfrac{1}{x^2}-16=\left(2x-\dfrac{1}{x}\right)^2-16=\left(2x-\dfrac{1}{x}-4\right)\left(2x-\dfrac{1}{x}+4\right)=\left(\dfrac{2x^2-1}{x}-4\right)\left(\dfrac{2x^2-1}{x}+4\right)\)