Cho A=\(4x^2-5x+\frac{1}{2}\)
B=\(-4x^2+2x-\frac{3}{2}\)
Tính A+B;A-B
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
8 tháng 10 2016
Ta có:
x = \(\frac{1}{2}\)\(\sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}}\)
= \(\frac{1}{2}\)\(\sqrt{\frac{\left(\sqrt{2}-1\right)^2}{1}}\)
= \(\frac{1}{2}\)(\(\sqrt{2}\)-1)
=> 2x = \(\sqrt{2}\)-1
=> (2x)2= ( \(\sqrt{2}\)-1)2
=> 4x2= 2-2\(\sqrt{2}\)+1
=> 4x2= -2( \(\sqrt{2}\)-1)+1
=> 4x2= -4x +1 => 4x2+4x-1=0
Lại có:
A1= (\(4x^5\)+\(4x^4\)- \(x^3\)+1)19
= [ x3( 4x2+4x-1) +1]19
=1
A2=( \(\sqrt{4x^5+4x^4-5x^3+5x+3}\))3
= (\(\sqrt{x^3\left(4x^2+4x-1\right)-x\left(4x^2+4x-1\right)+\left(4x^2+4x-1\right)+4}\))3
= 23=8
A3= \(\frac{1-\sqrt{2x}}{\sqrt{2x^2+2x}}\)
= \(\sqrt{2}\)- \(\sqrt{2}\)\(\sqrt{1-\sqrt{2}}\)
Cộng 3 số vào ta được A
28 tháng 3 2020
ĐKXĐ bạn tự tìm nha : )
k, Ta có : \(\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}=\frac{\left(1-2x\right)\left(1+2x\right)}{x\left(x+4\right)}.\frac{3x}{2\left(1-2x\right)}\)
\(=\frac{3x\left(1-2x\right)\left(1+2x\right)}{2x\left(x+4\right)\left(1-2x\right)}=\frac{3\left(1+2x\right)}{2\left(x+4\right)}\)
j, Ta có : \(\frac{x+y}{y-x}:\frac{x^2+xy}{3x^2-3y^2}=\frac{x+y}{y-x}:\frac{x\left(x+y\right)}{3\left(x^2-y^2\right)}=\frac{x+y}{y-x}.\frac{3\left(x-y\right)\left(x+y\right)}{x\left(x+y\right)}\)
\(=\frac{3\left(x-y\right)\left(x+y\right)}{x\left(y-x\right)}=\frac{3\left(x-y\right)\left(x+y\right)}{-x\left(x-y\right)}=\frac{-3\left(x+y\right)}{x}\)
i, Ta có : \(\frac{a^2+ab}{b-a}:\frac{a+b}{2a^2-2b^2}=\frac{a\left(a+b\right)}{-\left(a-b\right)}:\frac{a+b}{2\left(a^2-b^2\right)}=\frac{a\left(a+b\right)}{-\left(a-b\right)}.\frac{2\left(a-b\right)\left(a+b\right)}{a+b}\)
\(=\frac{2a\left(a+b\right)\left(a-b\right)}{-\left(a-b\right)}=-2a\left(a+b\right)\)
h, = k,
f, Ta có : \(\frac{x^2-36}{2x+10}.\frac{3}{6-x}=\frac{\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)}.\frac{-3}{x-6}=\frac{-3\left(x-6\right)\left(x+6\right)}{2\left(x+5\right)\left(x-6\right)}=\frac{-3\left(x+6\right)}{2\left(x+5\right)}\)
27 tháng 3 2020
a) Ta có: \(\frac{x+1}{2x+6}+\frac{2x+3}{x^2+3x}\)
\(=\frac{x+1}{2\left(x+3\right)}+\frac{2x+3}{x\left(x+3\right)}\)
\(=\frac{x\left(x+1\right)}{2x\left(x+3\right)}+\frac{2\cdot\left(2x+3\right)}{2x\left(x+3\right)}\)
\(=\frac{x^2+x+4x+6}{2x\left(x+3\right)}\)
\(=\frac{x^2+5x+6}{2x\left(x+3\right)}\)
\(=\frac{x^2+2x+3x+6}{2x\left(x+3\right)}\)
\(=\frac{x\left(x+2\right)+3\left(x+2\right)}{2x\left(x+3\right)}\)
\(=\frac{\left(x+2\right)\left(x+3\right)}{2x\left(x+3\right)}=\frac{x+2}{2x}\)
b) Ta có: \(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}\)
\(=\frac{3}{2x+6}-\frac{x-6}{x\left(2x+6\right)}\)
\(=\frac{3x}{x\left(2x+6\right)}-\frac{x-6}{x\left(2x+6\right)}\)
\(=\frac{3x-x+6}{x\left(2x+6\right)}=\frac{2x+6}{x\left(2x+6\right)}=\frac{1}{x}\)
c) Ta có: \(\frac{5x+10}{4x-8}\cdot\frac{4-2x}{x+2}\)
\(=\frac{5\left(x+2\right)\cdot2\cdot\left(2-x\right)}{4\cdot\left(x-2\right)\cdot\left(x+2\right)}\)
\(=\frac{5\cdot2\cdot\left(2-x\right)}{-4\left(2-x\right)}=\frac{5\cdot2}{-4}=\frac{-5}{2}\)
d) Ta có: \(\frac{1-4x^2}{x^2+4x}:\frac{2-4x}{3x}\)
\(=\frac{\left(1-2x\right)\left(1+2x\right)\cdot3x}{x\left(x+4\right)\cdot2\left(2-x\right)}\)
\(=\frac{\left(1-2x\right)\left(1+2x\right)\cdot3}{2\left(x+4\right)\cdot\left(2-x\right)}=\frac{3\left(1-4x^2\right)}{2\left(-x^2-2x+8\right)}\)
\(=\frac{3-12x^2}{-2x^2-4x+16}\)
JS
27 tháng 3 2020
a) \(\frac{x+1}{2x+6}+\frac{2x+3}{x^2+3x}\)
\(=\frac{x+1}{2\left(x+3\right)}+\frac{2x+3}{x\left(x+3\right)}\) \(\left(ĐKXĐ:x\ne-3;x\ne0\right)\)
\(=\frac{x^2+x}{2x\left(x+3\right)}+\frac{4x+6}{2x\left(x+3\right)}\)
\(=\frac{x^2+2x+3x+6}{2x\left(x+3\right)}=\frac{\left(x+2\right)\left(x+3\right)}{2x\left(x+3\right)}=\frac{x+2}{2x}\)
b) \(\frac{3}{2x+6}-\frac{x-6}{2x^2+6x}=\frac{3}{2\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}\) \(\left(ĐKXĐ:x\ne0;x\ne-3\right)\)
\(=\frac{3x}{2x\left(x+3\right)}-\frac{x-6}{2x\left(x+3\right)}=\frac{2\left(x+3\right)}{2x\left(x+3\right)}=\frac{1}{x}\)
c) \(\frac{5x+10}{4x-8}.\frac{4-2x}{x+2}=\frac{5\left(x+2\right)}{4\left(x-2\right)}.\frac{2\left(2-x\right)}{x+2}\) \(\left(ĐKXĐ:x\ne\pm2\right)\)
\(=\frac{-5\left(x-2\right)}{2\left(x-2\right)}=\frac{-5}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 3 2020
Bài 1:
\(a=\lim\limits_{x\rightarrow-\infty}\frac{2\left|x\right|+1}{3x-1}=\lim\limits_{x\rightarrow-\infty}\frac{-2x+1}{3x-1}=\lim\limits_{x\rightarrow-\infty}\frac{-2+\frac{1}{x}}{3-\frac{1}{x}}=-\frac{2}{3}\)
\(b=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{9+\frac{1}{x}+\frac{1}{x^2}}-\sqrt{4+\frac{2}{x}+\frac{1}{x^2}}}{1+\frac{1}{x}}=\frac{\sqrt{9}-\sqrt{4}}{1}=1\)
\(c=\lim\limits_{x\rightarrow+\infty}\frac{\sqrt{1+\frac{2}{x}+\frac{3}{x^2}}+4+\frac{1}{x}}{\sqrt{4+\frac{1}{x^2}}+\frac{2}{x}-1}=\frac{1+4}{\sqrt{4}-1}=5\)
\(d=\lim\limits_{x\rightarrow+\infty}\frac{\frac{3}{x}-\frac{2}{x\sqrt{x}}+\sqrt{1-\frac{5}{x^3}}}{2+\frac{4}{x}-\frac{5}{x^2}}=\frac{1}{2}\)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 3 2020
Bài 2:
\(a=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{1}{x}}{1-\frac{1}{x}}=2\)
\(b=\lim\limits_{x\rightarrow-\infty}\frac{2+\frac{3}{x^3}}{1-\frac{2}{x}+\frac{1}{x^3}}=2\)
\(c=\lim\limits_{x\rightarrow+\infty}\frac{x^2\left(3+\frac{1}{x^2}\right)x\left(5+\frac{3}{x}\right)}{x^3\left(2-\frac{1}{x^3}\right)x\left(1+\frac{4}{x}\right)}=\frac{15}{+\infty}=0\)
A+B=4x2-5x+1/2-4x2+2x-3/2=-3x-1
A-B=4x2-5x+1/2+4x2-2x+3/2=8x2-7x+2