Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a/ \(\dfrac{x^3}{x^2+1975}\cdot\dfrac{2x+1954}{x+1}+\dfrac{x^3}{x^2+1975}\cdot\dfrac{21-x}{x+1}=\dfrac{x^3\left(2x+1954\right)+x^3\left(21-x\right)}{\left(x^2+1975\right)\left(x+1\right)}=\dfrac{2x^4+1954x^3+21x^3-x^4}{\left(x^2+1975\right)\left(x+1\right)}=\dfrac{x^4+1975x^3}{\left(x^2+1975\right)\left(x+1\right)}\)
b/ \(\dfrac{19x+8}{x-7}\cdot\dfrac{5x-9}{x+1945}+\dfrac{19x+8}{x^2+1945}\cdot\dfrac{x-2}{x-7}=\dfrac{\left(19x+8\right)\left(5x-9\right)+\left(19x+8\right)\left(x-2\right)}{\left(x-7\right)\left(x+1945\right)}=\dfrac{\left(19x+8\right)\left(5x-9+x-2\right)}{\left(x-7\right)\left(x+1945\right)}=\dfrac{114x^2-209x+40x-88}{\left(x-7\right)\left(x+1945\right)}=\dfrac{114x^2-169x-88}{x^2+1938x-13615}\)
c/ \(\dfrac{x+1}{x^2-2x-8}\cdot\dfrac{4-x}{x^2+x}=\dfrac{\left(x+1\right)\left(4-x\right)}{x\left[x^2-4x+2x-8\right]\left(x+1\right)}=-\dfrac{x-4}{x\left(x-4\right)+2\left(x-4\right)}=-\dfrac{x-4}{\left(x-4\right)\left(x+2\right)}=-\dfrac{1}{x+2}\)
a) \(A=\left(3x-2\right)^2+\left(x+1\right)^2-2\left(x+1\right)\left(3x-2\right)\)
\(\Leftrightarrow A=\left(x+1\right)^2-2\left(x+1\right)\left(3x-2\right)+\left(3x-2\right)^2\)
\(\Leftrightarrow A=\left[\left(x+1\right)-\left(3x-2\right)\right]^2\)
\(\Leftrightarrow A=\left(x+1-3x+2\right)^2\)
\(\Leftrightarrow A=\left(3-2x\right)^2\)
Thay \(x=\dfrac{3}{2}\) vào biểu thức A ta được:
\(\left(3-2.\dfrac{3}{2}\right)^2=\left(3-3\right)^2=0^2=0\)
Vậy giá trị của biểu thức A tại \(x=\dfrac{3}{2}\) là 0
b) \(B=\dfrac{x^2y\left(y-x\right)-xy^2\left(x-y\right)}{3y^2-3x^2}\)
\(\Leftrightarrow B=\dfrac{x^2y\left(y-x\right)+xy^2\left(y-x\right)}{3\left(y^2-x^2\right)}\)
\(\Leftrightarrow B=\dfrac{\left(y-x\right)\left(x^2y+xy^2\right)}{3\left(y-x\right)\left(y+x\right)}\)
\(\Leftrightarrow B=\dfrac{xy\left(y-x\right)\left(x+y\right)}{3\left(y-x\right)\left(y+x\right)}\)
\(\Leftrightarrow B=\dfrac{xy\left(y-x\right)\left(y+x\right)}{3\left(y-x\right)\left(y+x\right)}\)
\(\Leftrightarrow B=\dfrac{xy}{3}\)
Thay \(x=-3\) và \(y=\dfrac{1}{2}\) vào biểu thức B ta được:
\(\dfrac{\left(-3\right).\dfrac{1}{2}}{3}=\dfrac{\dfrac{-3}{2}}{3}=\dfrac{\dfrac{-3}{2}}{3}=\dfrac{-1}{2}\)
Vậy giá trị của biểu thức B tại \(x=-3\) và \(y=\dfrac{1}{2}\) là \(\dfrac{-1}{2}\)
c) \(C=\dfrac{x+1}{x-3}-\dfrac{1-x}{x+3}-\dfrac{2x\left(1-x\right)}{9-x^2}\)
\(\Leftrightarrow C=\dfrac{x+1}{x-3}-\dfrac{1-x}{x+3}+\dfrac{2x\left(1-x\right)}{x^2-9}\)
\(\Leftrightarrow C=\dfrac{x+1}{x-3}-\dfrac{1-x}{x+3}+\dfrac{2x\left(1-x\right)}{\left(x-3\right)\left(x+3\right)}\) MTC: \(\left(x-3\right)\left(x+3\right)\)
\(\Leftrightarrow C=\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(x-3\right)\left(1-x\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{2x\left(1-x\right)}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow C=\dfrac{\left(x+1\right)\left(x+3\right)-\left(x-3\right)\left(1-x\right)+2x\left(1-x\right)}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow C=\dfrac{\left(x^2+3x+x+3\right)-\left(x-x^2-3+3x\right)+\left(2x-2x^2\right)}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow C=\dfrac{x^2+3x+x+3-x+x^2+3-3x+2x-2x^2}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow C=\dfrac{2x+6}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow C=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow C=\dfrac{2}{x-3}\)
Thay \(x=5\) vào biểu thức C ta được:
\(\dfrac{2}{5-3}=\dfrac{2}{2}=1\)
Vậy giá trị của biểu thức C tại \(x=5\) là 1
a: \(\dfrac{2x-3}{35}+\dfrac{x\left(x-2\right)}{7}\le\dfrac{x^2}{7}-\dfrac{2x-3}{5}\)
\(\Leftrightarrow2x-3+5x\left(x-2\right)\le5x^2-7\left(2x-3\right)\)
\(\Leftrightarrow2x-3+5x^2-10x< =5x^2-14x+21\)
=>-8x-3<=-14x+21
=>6x<=24
hay x<=4
b: \(\dfrac{6x+1}{18}+\dfrac{x+3}{12}>=\dfrac{5x+3}{6}+\dfrac{12-5x}{9}\)
=>2(6x+1)+3(x+3)>=6(5x+3)+4(12-5x)
=>12x+2+3x+9>=30x+18+48-20x
=>15x+11>=10x+66
=>5x>=55
hay x>=11
Câu trả lời sai là:
(C) Giá trị của Q tại \(x=3\) là \(\dfrac{3-3}{3+3}=0\)
Do ĐKXĐ của phương trình
\(Q=\dfrac{x^2-6x+9}{x^2-9}\) là \(x\ne\pm3\)
a: \(\Leftrightarrow5\left(x+1\right)\left(x-1\right)=2x-2-3x-3=-x-5\)
\(\Leftrightarrow5x^2-5+x+5=0\)
=>x(5x+1)=0
=>x=0 hoặc x=-1/5
b: \(\Leftrightarrow x^2-x-\left(2x-3\right)\left(x+1\right)=2x+3\)
\(\Leftrightarrow x^2-x-2x^2-2x+3x+3=2x+3\)
\(\Leftrightarrow-x^2+3=2x+3\)
=>-x(x+2)=0
=>x=0(nhận) hoặc x=-2(nhận)
c: \(\Leftrightarrow4x^2-25=0\)
=>(2x-5)(2x+5)=0
=>x=5/2 hoặc x=-5/2
1/ a, \(A=\dfrac{3}{2x+6}-\dfrac{x-6}{2x^2+6x}\)
\(=\dfrac{3}{2\left(x+3\right)}-\dfrac{x-6}{2x\left(x+3\right)}\)
\(=\dfrac{3x-x+6}{2x\left(x+3\right)}\)
\(=\dfrac{2x+6}{2x\left(x+3\right)}\)
\(=\dfrac{2\left(x+3\right)}{2x\left(x+3\right)}\)
\(=\dfrac{1}{x}\)
Vậy \(A=x\)
b/ Khi \(x=\dfrac{1}{2}\Leftrightarrow A=\dfrac{1}{\dfrac{1}{2}}=2\)
Vậy...
2/a,
\(A=\dfrac{5x+2}{3x^2+2x}+\dfrac{-2}{3x+2}\)
\(=\dfrac{5x+2}{x\left(3x+2\right)}-\dfrac{2x}{x\left(3x+2\right)}\)
\(=\dfrac{5x+2-2x}{x\left(3x+2\right)}\)
\(=\dfrac{3x+2}{x\left(3x+2\right)}\)
\(=\dfrac{1}{x}\)
Vậy....
b/ Với \(x=\dfrac{1}{3}\Leftrightarrow A=\dfrac{1}{\dfrac{1}{3}}=3\)
Vậy..
\(A=\dfrac{x^3}{x+1975}.\dfrac{2x+1954}{x+1}+\dfrac{x^3}{x+1075}.\dfrac{21-x}{x+1}\)
\(=\dfrac{x^3}{x+1975}\left(\dfrac{2x+1954}{x+1}+\dfrac{21-x}{x+1}\right)\)
\(=\dfrac{x^3}{x+1975}.\dfrac{2x+1954+21-x}{x+1}\)
\(=\dfrac{x^3}{x+1975}.\dfrac{x+1975}{x+1}\)
\(=\dfrac{x^3}{x+1}\)
Thay x = 3 vào biểu thức A ,có :
\(\dfrac{3^3}{3+1}=\dfrac{27}{4}\)
Vậy tại x = 3 giá trị cảu biểu thức A là \(\dfrac{27}{4}\)