tinh gia tri bieu thuc (1+1/2)x(1+1/3)x(1+1/4)x...x(1+1/2013)
Cho bieu thuc A = ( 1/ x^2 - x + 1/x-1):x+1/x^2 -2x +1 ( x khac 0;1;-1)
a) Rut gon bieu thuc A
b) Tinh gia tri bieu thuc A khi x=2014/2013
c)Tim dieu kien cua x de A co gia tri lon hon 1
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cho x+y =1 . tinh gia tri cua bieu thuc A=x^3+y^3+3xy
chox-y=1. tinh gia tri cua bieu thuc B=x^3-y^3-3xy
cho x+y=1 . tinh gia tri cua bieu thuc C=x^3+y^3+3xy(x^2+y^2)+6x^2*y^2(x+y)
Câu 1: Ta có: A = \(x^3+y^3+3xy=x^3+y^3+3xy\times1=x^3+y^3+3xy\left(x+y\right)\)
\(=\left(x+y\right)^3=1^3=1\)
Câu 2: Ta có: \(B=x^3-y^3-3xy=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\)
\(=x^2+xy+y^2-3xy=x^2-2xy+y^2=\left(x-y\right)^2=1^2=1\)
Câu 3: Ta có: \(C=x^3+y^3+3xy\left(x^2+y^2\right)-6x^2.y^2\left(x+y\right)\)
\(=x^3+y^3+3xy\left(x^2+2xy+y^2-2xy\right)+6x^2y^2\)
\(=x^3+y^3+3xy\left(x+y\right)^2-3xy.2xy+6x^2y^2\)
\(=x^3+y^3+3xy.1-6x^2y^2+6x^2y^3\)
\(=x^3+y^3+3xy\left(x+y\right)=\left(x+y\right)^3=1^3=1\)
1) Cho bieu thuc A=\(3+\frac{2}{x-1}\). Tinh gia tri cua bieu thuc A khi |2x-3|=1
2) Rut gon bieu thuc B=\(\frac{x}{x-1}\)-\(\frac{x-5}{x+1}\)-\(\frac{3-x}{1-x^2}\)
3) Tim cac gia tri nguyen cua x de bieu thuc \(\frac{B}{A}\)co gia tri nguyen duong
x(x+4)-6(x-1)(x+1)+(2x-1)2 rut gon bieu thuc a tinh gia tri cua x de a co gia tri bang 3
\(A=x\left(x+4\right)-6\left(x-1\right)\left(x+1\right)+\left(2x-1\right)^2\)
\(A=x^2+4x-6\left(x^2-1\right)+\left(4x^2-4x+1\right)\)
\(A=x^2+4x-6x^2+6+4x^2-4x+1\)
\(A=-x^2+7\)
Để A có giá trị bằng 3 thì :
\(-x^2+7=3\)
\(-x^2=-4\)
\(x^2=4\)
\(x\in\left\{\pm2\right\}\)
Vậy..........
tinh gia tri bieu thuc (x-1)^3-4x(x+1)(x-1)+3(x-1)(x^2+x+1) voi x=-2
Đặt \(A\) , ta có :
\(A=\left(x-1\right)^3-4x\left(x+1\right)\left(x-1\right)+3\left(x-1\right)\left(x^2+x+1\right)\)
\(A=\left(x-1\right)^3-4x.\left(x^2-1^2\right)+3.\left(x^3-1\right)\)
Thay \(x=2\) vào biểu thức , ta có :
\(A=\left(-2-1\right)^3-4.\left(-2\right).\left[\left(-2\right)^2-1\right]+3.\left[\left(-2\right)^3-1\right]\)
\(A=\left(-3\right)^3+8.3+3.\left(-9\right)\)
\(A=-27+24-27\)
\(A=-30\)
tinh gia tri bieu thuc (x-1)^3-4x(x+1)(x-1)+3(x-1)(x^2+x+1) voi x=-2
\(A=\left(x-1\right)^3-4x.\left(x+1\right).\left(x-1\right)+3.\left(x-1\right).\left(x^2+x+1\right)\)
\(A=\left(x-1\right)^3-4x.\left(x^2-1^2\right)+3.\left(x^3-1\right)\)
Thay x=2 vào biểu thức ta có
\(A=\left(-2-1\right)^3-4.\left(-2\right).\left[\left(-2\right)^2-1\right]+3.\left[\left(-2\right)^3-1\right]\)
\(A=\left(-3\right)^3+8.3+3.\left(-9\right)\)
\(A=-27+24-27\)
\(A=-30\)
a) Tinh gia tri bieu thuc:
3/5 : 4/5 + 1/2 x 2/3
b) Tim x:
5/4 nhan x = 3/8 + 1/4
A
\(\frac{3}{5}\div\frac{4}{5}+\frac{1}{2}\times\frac{2}{3}\)
\(=\frac{3}{4}+\frac{1}{3}\)
\(=\frac{13}{12}\)
B
\(\frac{5}{4}\times x=\frac{3}{8}+\frac{1}{4}\)
\(\frac{5}{4}\times x =\frac{5}{8}\)
\(x=\frac{5}{8}\div\frac{5}{4}\)
\(x=\frac{4}{8}=\frac{1}{2}\)
cho 2 bieu thuc A=x+x^2/2-x va B=2x/x+1+3/x-2-2x^2+1/x^2-x-2 a, tinh gia tri cua A khi /2x-3/=1 b,tim dieu kien xac dinh va rut gon bieu thuc B c,tim so nguyen x de P=A.B dat gia tri lon nhat
mk dang can gap
a:
ĐKXĐ: x<>2
|2x-3|=1
=>\(\left[{}\begin{matrix}2x-3=1\\2x-3=-1\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=2\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)
Thay x=1 vào A, ta được:
\(A=\dfrac{1+1^2}{2-1}=\dfrac{2}{1}=2\)
b: ĐKXĐ: \(x\notin\left\{-1;2\right\}\)
\(B=\dfrac{2x}{x+1}+\dfrac{3}{x-2}-\dfrac{2x^2+1}{x^2-x-2}\)
\(=\dfrac{2x}{x+1}+\dfrac{3}{x-2}-\dfrac{2x^2+1}{\left(x-2\right)\left(x+1\right)}\)
\(=\dfrac{2x\left(x-2\right)+3\left(x+1\right)-2x^2-1}{\left(x+1\right)\left(x-2\right)}\)
\(=\dfrac{2x^2-4x+3x+3-2x^2-1}{\left(x+1\right)\left(x-2\right)}\)
\(=\dfrac{-x+2}{\left(x+1\right)\left(x-2\right)}=-\dfrac{1}{x+1}\)
c: \(P=A\cdot B=\dfrac{-1}{x+1}\cdot\dfrac{x\left(x+1\right)}{2-x}=\dfrac{x}{x-2}\)
\(=\dfrac{x-2+2}{x-2}=1+\dfrac{2}{x-2}\)
Để P lớn nhất thì \(\dfrac{2}{x-2}\) max
=>x-2=1
=>x=3(nhận)
tinh gia tri bieu thuc
1/2x+2/3(x-1)=1/3
Ta có:\(\frac{1}{2x}+\frac{2}{3\left(x-1\right)}=\frac{1}{3}\)
\(\frac{3\left(x-1\right)}{6x\left(x-1\right)}+\frac{4x}{6x\left(x-1\right)}=\frac{1}{3}\)
\(\frac{3x-3+4x}{6x\left(x-1\right)}=\frac{1}{3}\)
\(\frac{7x-3}{6x\left(x-1\right)}=\frac{1}{3}\)
\(\Rightarrow21x-9=6x^2-6x\)
\(\Rightarrow21x-9-6x^2+6x=0\)
\(\Rightarrow-6x^2+27x-9=0\)
Đến đây mk gợi ý thôi nha