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
I> Cho bieu thuc
A = ( \(\frac{2x+1}{2x-1}\)- \(\frac{2x-1}{2x+1}\)) : \(\frac{8x}{3-6x}\)( voi x khac + - \(\frac{1}{2}\)x khax 0 )
a . Rut gon bieu thuc A
b . Tim x de A = \(\frac{3}{-4031}\)
II> Cho bieu thuc
B = ( \(\frac{1}{1-x}\)+\(\frac{2}{x+1}\)-\(\frac{5-x}{1-x^2}\)) : \(\frac{1-2x}{x^2-4}\)
a . Rut gon bieu thuc B
b . Tìm giá trị nguyên của x để giá trị của biểu thức B là số nguyên
Chung minh bieu thuc Q=(x^4*y^n+1-1/2*x^3*y^n+2):1/2x^3*y^n-20x^4*y:5*xy^2 (n thuoc N) luon <0 voi moi gia tri x khac 0,y khac 0
rut gon bieu thuc M=1/x - 2/(5-x) - x+5/(x^2-5x) voi x khac 0 ; x khac 5
a, Với x ≠ 0,x ≠ ± 5 và x ≠ 5/2 thì
P = [x/(x^2 - 25) - (x - 5)/(x^2 + 5x)] : (2x - 5)/(x^2 + 5x) + x/(x - 5)
<=>P = [x/(x - 5)(x + 5) - (x - 5)/x(x+5)] . x(x + 5)/(2x - 5) + x/(x - 5)
=> P = [x^2 - (x - 5)^2]/x(x - 5)(x + 5) . x(x + 5)/(2x - 5) + x/(x - 5)
<=> P = (x - x + 5)(x + x - 5)/(x - 5)(2x - 5) + x/(x - 5)
<=> P = 5(2x - 5)/(x - 5)(2x - 5) + x/(x - 5)
<=> P = 5/(x - 5) + x/(x - 5)
<=> P = (5 + x)/(x - 5)
b, Với x ≠ 0,x ≠ ± 5 và x ≠ 5/2 (x ∈ Z) thì P ∈ Z <=> (5 + x)/(x - 5) ∈ Z
<=> (x - 5 + 10)/(x - 5) ∈ Z
<=> 1 + 10/(x - 5) ∈ Z
<=> 10/(x - 5) ∈ Z
<=> (x - 5) ∈ Ư(10)
<=> x - 5 = 10 <=> x = 15 (TM)
hoặc x - 5 = -10 <=> x = -5 (TM)
hoặc x - 5 = 5 <=> x = 10 (TM)
hoặc x - 5 = -5 <=> x = 0 (TM)
hoặc x - 5 = 2 <=> x = 7 (TM)
hoặc x - 5 = -2 <=> x = 3 (TM)
hoặc x - 5 = -1 <=> x = 4 (TM)
hoặc x - 5 = 1 <=> x = 6 (TM)
Vậy x ∈ {-5,0,3,4,6,7,10,15} thì P ∈ Z
cho 2 bieu thuc:
A=(\(\sqrt{20}\) -\(\sqrt{45}\) +3\(\sqrt{5}\) ).\(\sqrt{5}\) va B=\(\dfrac{x+1-2\sqrt{x}}{\sqrt{x}-1}\) +\(\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\) (Dieu kien: x>0, x khac 1
a) Rut gon bieu thuc A va B
b)Tim cac gia tri cua x de gia tri cua bieu thuc A bang 2lan gia tri B
a: \(A=\left(2\sqrt{5}-3\sqrt{5}+3\sqrt{5}\right)\cdot\sqrt{5}=2\sqrt{5}\cdot\sqrt{5}=10\)
\(B=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(=\sqrt{x}-1+\sqrt{x}=2\sqrt{x}-1\)
b: A=2B
=>\(10=4\sqrt{x}-2\)
=>\(4\sqrt{x}=12\)
=>x=9(nhận)
Cho x,y,z khac 0 va x - y -z = 0. Tinh gia tri bieu thuc A = ( 1- z/x)(1-x/y)(1-y/z)
Cho bieu thuc A=x^4-x^3+2x^2-x+1/x^4-x^3+2x^2-x+1 . Chung minh bieu thuc A khong am .
Cho x,y,z khac o va x-y-z=0.Tinh gia tri cua bieu thuc A=(1-z/x)(1-x/y)(1+y/Z)
cho x,y,z đôi một khac nhau va 1/x +1/y+1/z=0 .tinh gia tri bieu thuc A=(x.y/x2+2yz)+(xz/y2+2xz)+(xy/z2+2xy)
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)