A=(2x-3)2-(x-3)3+(4x+1)(16x2-4x+1) voi x=-2 rut gon va tinh gia tri
lam kieu dang toan 8 ho minh nha
Những câu hỏi liên quan
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)
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chung minh rang 4x2- 4x+ 10>0 voi moi so thuc x
rut gon
x^2 - 4x+ 4
-2x + x^2
va tinh gia tri cua A khi x = -1/2
b) phan tich da thuc thanh nhan tu
a^3 + 8a- 4a - 8
giup minh nhe
Rut gon a biet a=(x^2-2x/2^2+8+2x^2/8-4x+2x^2-x^3)x(1-1/-2/^2)
Ai giup minh minh tick nha ::)))
M=(x2-2x/2x2+8 - 2x2/8-4x+2x2-x3)*(1-1/x-2/x2)
a, Rut gon M
b,Tinh M voi x=1/2
c,tinh x de M<0
cho P=(2+x/2-x+4x^2/x^2-4-2-x/2+x):x^2-3x/2x^2-x^3
a) tim dieu kien cua x de gia tri cua P xac dinh
b) rut gon P
mn giup minh voi
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..........
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Rut gon va tinh gia tri bieu thuc
A= (x-4).(x-2)-(x-1).(x-3) tại x= \(1\dfrac{3}{4}\)
Ta có:
\(A=\left(x-4\right)\left(x-2\right)-\left(x-1\right)\left(x-3\right)\)
\(A=\left(x^2-4x-2x+8\right)-\left(x^2-x-3x+4\right)\)
\(A=\left(x^2-6x+8\right)-\left(x^2-4x+4\right)\)
\(A=x^2-6x+8-x^2+4x-4\)
\(A=-2x+4\)
Thay \(x=1\dfrac{3}{4}=\dfrac{7}{4}\) vào A ta được:
\(A=-2.\dfrac{7}{4}+4\)
\(A=-\dfrac{7}{2}+4\)
\(A=\dfrac{1}{2}\)
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Bai1:
a- 3x^2 - 7x +2 b- a(x^2 +1 ) - x (a^2 +1)
Bai2 : Cho bieu thuc
A= (2+x/ 2-x - 4x^2 / x^2 -4 - 2-x/ 2+x ) : x^2 - 3x / 2x^2 - x ^3
a- Tim DKXD roi rut gon bieu thuc A ?
b- Tim gia tri cua x de A>0?
c- Tinh gia tri cua A TRong truong hop : \vbar x-7 \vbar=4
Giup mk nha moi nguoi !!!!!!!!!
Bài 2:
a: ĐKXĐ: \(x\notin\left\{0;2;-2;3\right\}\)\(A=\left(\dfrac{-\left(x+2\right)}{x-2}-\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{x+2}\right):\dfrac{x\left(x-3\right)}{x^2\left(2-x\right)}\)
\(=\dfrac{-x^2-4x-4-4x^2+x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-x\left(x-2\right)}{x-3}\)
\(=\dfrac{-4x^2-8x}{\left(x+2\right)}\cdot\dfrac{-x}{x-3}\)
\(=\dfrac{-4x\left(x+2\right)}{x+2}\cdot\dfrac{-x}{x-3}=\dfrac{4x^2}{x-3}\)
b: Để A>0 thì x-3>0
hay x>3
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Thu gon roi tinh gia tri tai x = 1, y = \(-\dfrac{1}{2}\)
M = ( 5x - 3y + 3xy + x2y2) - (\(\dfrac{1}{2}\)x + 2xy - y + 4x2y2)
Mn giup em voi a ! Em cam on nhieu lam ! Em dang can gap vo cung moi nguoi nhich chut thoi gian giup em a ! Mon nao em cx hoc dc het chi co Toan la bo tay thoi ! Mn giup e cam on truoc a
\(M=\left(5x-3y+3xy+x^2y^2\right)-\left(\dfrac{1}{2}x+2xy-y+4x^2y^2\right)\)
\(=5x-3y+3xy+x^2y^2-\dfrac{1}{2}x-2xy+y-4x^2y^2\)
\(=\left(5x-\dfrac{1}{2}x\right)+\left(y-3y\right)+\left(3xy-2xy\right)+\left(x^2y^2-4x^2y^2\right)\) \(=4,5x-2y+xy-3x^2y^2\)
Thay \(x=1;y=-\dfrac{1}{2}\) vào ta có:
\(4,5x-2y+xy-3x^2y^2\)
\(=4,5.1-2.\left(-\dfrac{1}{2}\right)+1.\left(-\dfrac{1}{2}\right)-3.1^2.\left(-\dfrac{1}{2}\right)^2\)
\(=4,5+1-\dfrac{1}{2}-\dfrac{3}{4}\) \(=\dfrac{17}{4}\)
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