cho bieu thuc A = [ -a - b + c ] - [ -a - b -c ]
rut gon a . tih gi tri A khi a = 1 , b = -1 , c = -2
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
Mọi người tk mình đi mình đang bị âm nè!!!!!!
Ai tk mình mình tk lại nha !!!
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)
Cho phan thuc B=(3\y+3)+(1\y-3)-(18\9-y2)
a)Tim dieu kien cua y de gia tri cua bieu thuc B duoc xac dinh
b)Rut gon bieu thuc B
c)Tinh gia tri cua B de B co gia tri nguyen
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 \(1+3\sqrt{3}\).
giai dum minh bai nay voi :c=x/2x-2+x^2+1/2-2x^2
a)tim x de a co nghia
b)rut gon bieu thuc c
c)tim gia tri cua x de bieu thuc kia =1/2
giúp mình cảm ơn nhìu nha !
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
Cho bieu thuc:P=\(\dfrac{\left(a+3\right)^2}{a^2+3a}\times\left(1-\dfrac{6a-18}{a^2-9}\right)\)voi a ≠0;a≠ +-3
a)rut gon bieu thuc P
b)tim gia tri cua a de P= -2
c)tim cac gia tri nguyen cua a de bieu thuc P co gia tri nguyen
mng giup minh voi mai thi rui!
a: \(P=\dfrac{a+3}{a}\cdot\dfrac{a^2-9-6a+18}{\left(a-3\right)\left(a+3\right)}\)
\(=\dfrac{\left(a-3\right)^2}{a\left(a-3\right)}=\dfrac{a-3}{a}\)
b: Để P=-2 thì -2a=a-3
=>-3a=-3
=>a=1
c: Để P nguyên thì a-3 chia hết cho a
=>-3 chia hết cho a
mà a<>0; a<>3; a<>-3
nên \(a\in\left\{1;-1\right\}\)
B1: rut gon bieu thuc
a, (x+y)^2-4(x-y)^2
b, 2(x-y)(x+y)+(x+y)^2+(x-y)^2
B2: tim X
a, (2X-1)^2-4(X+2)^2=9
b, 3(X-1)^2-3X(X-5)=21
B3: Cho bieu thuc
M=(x+3)^3-(x-1)^3+12x(x-1)
a, Rut gon bieu thuc tren
b, Tinh gia tri M tai x=-2/3
c, Tim x de M=16
1)a)=>x2+y2+2xy-4(x2-y2-2xy)
=>x2+y2+2xy-4.x2+4y2+8xy
=>-3.x2+5y2+10xy
Cho bieu thuc: ( x-1/ x+1 - x-1/x+1) : 2x / 3x - 3
a, Tim dieu kien xac dinh cua bieu thuc P
b, Rut gon bieu thuc P
c, Tim x thuoc z de P nhan gia tri nguyen.
Đề bài sai rồi bạn ! Mình sửa :
a) \(ĐKXĐ:\hept{\begin{cases}x\ne0\\x\ne\pm1\end{cases}}\)
b) \(P=\left(\frac{x-1}{x+1}-\frac{x+1}{x-1}\right):\frac{2x}{3x-3}\)
\(\Leftrightarrow P=\frac{\left(x-1\right)^2-\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}\cdot\frac{3\left(x-1\right)}{2x}\)
\(\Leftrightarrow P=\frac{x^2-2x+1-x^2-2x-1}{\left(x-1\right)\left(x+1\right)}\cdot\frac{3\left(x-1\right)}{2x}\)
\(\Leftrightarrow P=\frac{-4x}{\left(x-1\right)\left(x+1\right)}\cdot\frac{3\left(x-1\right)}{2x}\)
\(\Leftrightarrow P=\frac{-6}{x+1}\)
c) Để P nhận giá trị nguyên
\(\Leftrightarrow\frac{-6}{x+1}\inℤ\)
\(\Leftrightarrow x+1\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
\(\Leftrightarrow x\in\left\{-2;0;-3;1;-4;2;-7;5\right\}\)
Ta loại các giá trị ktm
\(\Leftrightarrow x\in\left\{-2;-3;-4;2;-7;5\right\}\)
Vậy để \(P\inℤ\Leftrightarrow x\in\left\{-2;-3;-4;2;-7;5\right\}\)