rut gon bieu thuc sau (voi x<-1,3)
a) A= /x+1,3/ - /x-2,5/
b) B =/-x-1,3/+/x-2,5/
cho bieu thuc p=(x+1)(x+√x)/√x-x-√x, voi x>0
a/ rut gon bieu thuc
b/ tim gia tri cua x de gia tri cua bieu thuc p bang 2
Rut gon bieu thuc sau : A=3x5x7x11x13x37-10101 / 1212120+40404
giup minh voi cac ban oi!!!!
bn ơi ghi lại đề bài = phân số đc hk ghi cả nhân hay chia nx khó nhìn lắm!
A=\(\frac{3.5.7.11.13.37-10101}{1212120+40404}\)
A=\(\frac{\left(3.7.11.37.13\right).5-10101}{120.10101+4.10101}\)
A=\(\frac{10101.5-10101}{120.10101+4.10101}\)
A=\(\frac{1.5-1}{120.1+4.1}\)
A=\(\frac{5-1}{120+4}\)
A=\(\frac{4}{120}\)
A=\(\frac{1}{31}\)
# Học tốt
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
voi a<-1, gia tri rut gon cua bieu thuc /a/ +a la............
Cho bieu thuc:
P=\(\frac{1}{\sqrt{x}+2}-\frac{5}{x-\sqrt{x}-6}-\frac{\sqrt{x}-2}{3-\sqrt{x}}\)
a. Rut gon bieu thuc P
b.Tim GTLN cua P sau khi rut gon
đk: x>=0; x khác 3
a) \(P=\frac{\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}-\frac{5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}-3}=\frac{\sqrt{x}-3-5+x-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\frac{x+\sqrt{x}-12}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\)
\(P=\frac{\left(\sqrt{x}+4\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+4}{\sqrt{x}+2}\)
b) \(P=\frac{\sqrt{x}+2+2}{\sqrt{x}+2}=1+\frac{2}{\sqrt{x}+2}\)
ta có: \(x\ge0\Rightarrow\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+2\ge2\Leftrightarrow\frac{2}{\sqrt{x}+2}\le1\Leftrightarrow1+\frac{2}{\sqrt{x}+2}\le2\Rightarrow MaxP=2\Rightarrow x=0\)
rut gon bieu thuc sau:
(x+5)^3-15x(x+10)
rut gon bieu thuc 2x2\(\sqrt{\frac{9}{x^4}}\) voi x<0
Với x < 0, có:
\(2x^2\sqrt{\frac{9}{x^4}}\) = \(\frac{2x^2.9}{x^2}=18\)
rut gon bieu thuc sau -4+{x-15}+3
-4+(x-15)+3
=-4+x-15+3
=x-19+3
=x-16
nhớ k mk nha
cho bieu thuc:P=\(\frac{\sqrt{x}}{\sqrt{x}-3}\)+\(\frac{2\sqrt{x}}{\sqrt{x}-3}\)--\(\frac{3x+9}{x-9}\) voi x>= 0;x#9 .a; Rut gon bieu thuc P . b; Tinh gia tri cua bieu thuc voi \(x=4-2\sqrt{3}\)