1.Tinh gia tri cua bieu thuc
\(\dfrac{1}{2}+\dfrac{1}{3}\).x-\(\dfrac{1}{6}\).x voi x=\(\dfrac{-3}{5}\)
Cho bieu thuc \(P=\left(\dfrac{3}{x-1}-\dfrac{1}{\sqrt{x}+1}\right):\dfrac{1}{\sqrt{x}+1}\)
a.Neu dkxd va rut gon bieu thuc P
b.Tim cac gia tri cua x de \(P=\dfrac{5}{4}\)
c.Tim gia tri nho nhat cua bieu thuc :\(M=\dfrac{x+12}{\sqrt{x}-1}\cdot\dfrac{1}{P}\)
a)ĐKXĐ:x>0
P=\(\left(\frac{3}{x-1}-\frac{1}{\sqrt{x}+1}\right):\frac{1}{\sqrt{x}+1}\left(vớix>0\right)\)
=\(\left[\frac{3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}+1}\right]:\frac{1}{\sqrt{x}+1}\)
=\(\left[\frac{3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right]:\frac{1}{\sqrt{x}+1}\)
= \(\left[\frac{3-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]:\frac{1}{\sqrt{x}+1}\)
=\(\frac{4-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{\sqrt{x}+1}{1}\)
=\(\frac{4-\sqrt{x}}{\sqrt{x}-1}\)
b)Để P=\(\frac{5}{4}\left(vớix>0\right)\)
\(\Leftrightarrow\frac{4-\sqrt{x}}{\sqrt{x}-1}=\frac{5}{4}\)
\(\Leftrightarrow\frac{4-\sqrt{x}}{\sqrt{x}-1}-\frac{5}{4}=0\)
\(\Leftrightarrow\frac{4\left(4-\sqrt{x}\right)}{4\left(\sqrt{x}-1\right)}-\frac{5\left(\sqrt{x}-1\right)}{4\left(\sqrt{x}-1\right)}=0\)
\(\Rightarrow16-4\sqrt{x}-5\sqrt{x}+5=0\)
\(\Leftrightarrow21-9\sqrt{x}=0\)
\(\Leftrightarrow-9\sqrt{x}=-21\)
\(\Leftrightarrow\sqrt{x}=\frac{7}{3}\)
\(\Leftrightarrow x=\frac{21}{9}\)
Vậy:Để P=\(\frac{5}{4}\)thì x=\(\frac{21}{9}\)
c)Còn phần c thì mik chịu
Cho bieu thuc A=\(\left(\dfrac{4}{x-\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\div\dfrac{1}{\sqrt{x}-1}\)
a/ Tim dieu kien cua x de bieu thuc A co gia tri xac dinh
b/ Rut gon A
c/ Tinh gia tri cua A khi x = \(4-2\sqrt{3}\)
d/ Tim gia tri nho nhat cua A
a. ĐKXĐ : x>1.
b. \(A=\left(\dfrac{4}{x-\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right):\dfrac{1}{\sqrt{x}-1}=\left[\dfrac{4}{\sqrt{x}\left(\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right].\left(\sqrt{x}-1\right)=\dfrac{4+\sqrt{x}.\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\left(\sqrt{x}-1\right)=\dfrac{4+x}{\sqrt{x}}\)
c. Thay \(x=4-2\sqrt{3}\) vào A, ta có:
\(A=\dfrac{4+4-2\sqrt{3}}{\sqrt{4-2\sqrt{3}}}=\dfrac{8-2\sqrt{3}}{\sqrt{\left(\sqrt{3}-1\right)^2}}=\dfrac{8-2\sqrt{3}}{\sqrt{3}-1}=\dfrac{\left(8-2\sqrt{3}\right)\left(\sqrt{3}+1\right)}{3-1}=\dfrac{8\sqrt{3}+8-6-2\sqrt{3}}{2}=\dfrac{2+6\sqrt{3}}{2}=\dfrac{2\left(1+3\sqrt{3}\right)}{2}=1+3\sqrt{3}\)
Vậy giá trị của A tại \(x=4-2\sqrt{3}\) là \(1+3\sqrt{3}\).
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)
tinh gia tri cua bieu thuc :
\(\dfrac{x^3}{8}+\dfrac{x^2y}{4}+\dfrac{xy^2}{6}+\dfrac{y^3}{27}taix=-8;y=6\)
\(\dfrac{1}{3}x^8+\dfrac{1}{4}x^2y+\dfrac{1}{6}xy^2+\dfrac{1}{27}y^3\)
\(=\left(\dfrac{1}{2}x\right)^3+3\cdot\left(\dfrac{1}{2}x\right)^2\cdot\dfrac{1}{3}y+3\cdot\dfrac{1}{2}x\cdot\dfrac{1}{9}y^2+\left(\dfrac{1}{3}y\right)^3\)
\(=\left(\dfrac{1}{2}x+\dfrac{1}{3}y\right)^3\)
\(=\left(-4+2\right)^3=-8\)
Cho 2 bieu thuc :
A=\(\dfrac{x-3}{x+2}va\) B= \(\dfrac{3}{x+3}+\dfrac{2}{x-3}-\dfrac{3x-9}{x^2-9}\left(x-2,x\ne3x\ne-3\right)\)
a, Tinh gia tri bieu thuc A khi x=5
b, Chung minh : B=\(\dfrac{2}{x-3}\)
c, Biet C = A.B, Tim x de c = \(\dfrac{-1}{3}\)
\(a,A=\dfrac{5-3}{5+2}=\dfrac{2}{7}\\ b,B=\dfrac{3x-9+2x+6-3x+9}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ c,C=AB=\dfrac{x-3}{x+2}\cdot\dfrac{2}{x-3}=\dfrac{2}{x+2}\\ C=-\dfrac{1}{3}\Leftrightarrow x+2=-6\Leftrightarrow x=-8\left(tm\right)\)
Chung minh voi x\(\ne\)\(\pm\)1, gia tri cua bieu thuc sau khong phu thuoc vao gia tri cua bien:
A=(\(\dfrac{x+3}{2x+2}\)+\(\dfrac{3}{1-x^2}\)-\(\dfrac{x+1}{2x-2}\))\(\div\)\(\dfrac{3}{2x^2-2}\)
ta có:
A = \(\left(\dfrac{x+3}{2x+2}+\dfrac{3}{1-x^2}-\dfrac{x+1}{2x-2}\right):\dfrac{3}{2x^2-2}\)
= \(\left(\dfrac{x+3}{2\left(x+1\right)}-\dfrac{3}{x^2-1}-\dfrac{x+1}{2\left(x-1\right)}\right):\dfrac{3}{2\left(x^2-1\right)}\)
= \(\left(\dfrac{x+3}{2\left(x+1\right)}-\dfrac{3}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+1}{2\left(x-1\right)}\right):\dfrac{3}{2\left(x-1\right)\left(x+1\right)}\)
= \(\left(\dfrac{\left(x+3\right)\left(x-1\right)}{2\left(x+1\right)\left(x-1\right)}-\dfrac{6}{2\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}\right):\dfrac{3}{2\left(x-1\right)\left(x+1\right)}\)
= \(\left(\dfrac{x^2-x+3x-3-6-x^2-2x-1}{2\left(x+1\right)\left(x-1\right)}\right):\dfrac{3}{2\left(x-1\right)\left(x+1\right)}\)
= \(-\dfrac{10}{2\left(x+1\right)\left(x-1\right)}.\dfrac{2\left(x+1\right)\left(x-1\right)}{3}\)
= \(-\dfrac{10}{3}\)
Vậy phương trình trên ko phụ thuộc vào biến
cho x,y,z la cac so nguyen duong thoa man \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=2015\)
tinh gia tri lon nhat cua bieu thuc P=\(\dfrac{xy}{x^3+y^3}+\dfrac{yz}{y^3+z^3}+\dfrac{zx}{z^{3+x^3}}\)
cho bieu thuc
P=\(\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{1}{\sqrt{x}-2}\right).\dfrac{\sqrt{x}-2}{2}\)với x>=0,x≠4
a. tim gia tri cua P khi x=64
b. rút gọn bieu thuc p
c. tim cac gia tri cua x de bieu thuc 2P nhan gia tri nguyen
b \(P=\dfrac{\sqrt{x}+\sqrt{x}+2}{x-4}\cdot\dfrac{\sqrt{x}-2}{2}=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
a: Khi x=64 thì \(P=\dfrac{8+1}{8+2}=\dfrac{9}{10}\)
cho bieu thuc
P=\(\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{1}{\sqrt{x}-2}\right).\dfrac{\sqrt{x}-2}{2}\)với x>=0,x≠4
a. tim gia tri cua P khi x=64
b. rút gọn bieu thuc p
c. tim cac gia tri cua x de bieu thuc 2P nhan gia tri nguyen
b: \(P=\dfrac{\sqrt{x}+\sqrt{x}+2}{x-4}\cdot\dfrac{\sqrt{x}-2}{2}=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
a: Khi x=64 thì \(P=\dfrac{8+1}{8+2}=\dfrac{9}{10}\)