bai1 tinh gia tri cua bieu thuc
C=1,7+/3,4-x/
Những câu hỏi liên quan
bai1 :P=x+2/x+3-5/(x+3)(x-2)+1/2-x
a,Tim dkxd cua P
b,Rut gon bieu thuc P
c,tim x de P=-3/4
d,tim gia tri nguyen cua x de P cung co gia tri nguyen
e,tinh gia tri bieu thuc P khi x^2-9=0
cho x+y =1 . tinh gia tri cua bieu thuc A=x^3+y^3+3xy
chox-y=1. tinh gia tri cua bieu thuc B=x^3-y^3-3xy
cho x+y=1 . tinh gia tri cua bieu thuc C=x^3+y^3+3xy(x^2+y^2)+6x^2*y^2(x+y)
Câu 1: Ta có: A = \(x^3+y^3+3xy=x^3+y^3+3xy\times1=x^3+y^3+3xy\left(x+y\right)\)
\(=\left(x+y\right)^3=1^3=1\)
Câu 2: Ta có: \(B=x^3-y^3-3xy=\left(x-y\right)\left(x^2+xy+y^2\right)-3xy\)
\(=x^2+xy+y^2-3xy=x^2-2xy+y^2=\left(x-y\right)^2=1^2=1\)
Câu 3: Ta có: \(C=x^3+y^3+3xy\left(x^2+y^2\right)-6x^2.y^2\left(x+y\right)\)
\(=x^3+y^3+3xy\left(x^2+2xy+y^2-2xy\right)+6x^2y^2\)
\(=x^3+y^3+3xy\left(x+y\right)^2-3xy.2xy+6x^2y^2\)
\(=x^3+y^3+3xy.1-6x^2y^2+6x^2y^3\)
\(=x^3+y^3+3xy\left(x+y\right)=\left(x+y\right)^3=1^3=1\)
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gia tri nho nhat cua bieu thuc
-1,7+/x-2016/
\(\left|x-2016\right|\ge0\)
\(\Leftrightarrow-1,7+\left|x-2016\right|\ge-1,7\)
Dấu "=" xảy ra khi : \(\left|x+2016\right|=0\)
\(\Leftrightarrow x=-2016\)
Vậy ...
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bai 1 tinh gia tri cua bieu thuc 78 x m + 22 x m voi m =135
bai 2 tinh gia tri cua bieu thuc 78 x m + 42 x m -20 x m voi m =1035
bai 3 cho bieu thuc B = 119 x n - n x 9 bieu thuc B co gia tri bang 8470 khi n =............
cho bieu thuc A=[x+2/x^2-x+x-2/x^2+x].x^2-1/x^2+2
a) tim dieu kien cua x de gia tri cua bieu thuc A duoc xac dinh
b) tinh gia tri cua bieu thuc A voi x = -200
a) \(A=\left[\dfrac{x+2}{x^2-x}+\dfrac{x-2}{x^2+x}\right].\dfrac{x^2-1}{x^2-x}\)
\(A=\left[\dfrac{x+2}{x\left(x-1\right)}+\dfrac{x-2}{x\left(x+1\right)}\right].\dfrac{x^2-1}{x^2+2}\)
\(A=\left[\dfrac{\left(x+2\right)\left(x+1\right)+\left(x-2\right)\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}\right].\dfrac{x^2-1}{x^2+2}\)
\(A=\left[\dfrac{x^2+2x+x+2+x^2-2x-x+2}{x\left(x-1\right)\left(x+1\right)}\right].\dfrac{x^2-1}{x^2+2}\)
\(A=\dfrac{2x^2+4}{x\left(x^2-1\right)}.\dfrac{x^2-1}{x^2+2}\)
\(A=\dfrac{2\left(x^2+2\right)\left(x^2-1\right)}{x\left(x^2-1\right)\left(x^2+2\right)}=\dfrac{2}{x}\)
b) Thay \(x=-200\) vào biểu thức \(A=\dfrac{2}{x}\) ta được :
\(A=\dfrac{2}{x}=\dfrac{2}{-200}=\dfrac{-2}{200}=\dfrac{-1}{100}\)
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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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cho a R b =a*b/ a+b . Tinh gia tri bieu thuc cua : 13R7 - 12r4 . TINH gia tri cua bieu thuc tren ( 30 giay )
\(\frac{31}{20}\)đáp số đúng 1000000000000000000000000000000000000000000000000000% đó nha
h and kb nhé
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Xem thêm câu trả lời
cho pt: x^2-12x+4=0 c hai nghiem phan biet x1,x2. Khong giai pt, hay tinh gia tri cua bieu thuc: T=x1^2+x2^2/canx1+can x2cho pt: x^2-12x+4=0 c hai nghiem phan biet x1,x2. Khong giai pt, hay tinh gia tri cua bieu thuc: T=x1^2+x2^2/canx1+can x2
Ta có: \(\Delta'=32>0\)
\(\Rightarrow\) Phương trình có 2 nghiệm phân biệt
Theo Vi-ét, ta có: \(\left\{{}\begin{matrix}x_1+x_2=12\\x_1x_2=4\end{matrix}\right.\)
Mặt khác: \(T=\dfrac{x_1^2+x^2_2}{\sqrt{x_1}+\sqrt{x_2}}\)
\(\Rightarrow T^2=\dfrac{x_1^4+x^4_2+2x_1^2x_2^2}{x_1+x_2+2\sqrt{x_1x_2}}=\dfrac{\left(x_1^2+x_1^2\right)^2}{x_1+x_2+2\sqrt{x_1x_2}}\) \(=\dfrac{\left[\left(x_1+x_2\right)^2-2x_1x_2\right]^2}{x_1+x_2+2\sqrt{x_1x_2}}=\dfrac{\left(12^2-2\cdot4\right)^2}{12+2\sqrt{4}}=1156\)
Mà ta thấy \(T>0\) \(\Rightarrow T=\sqrt{1156}=34\)
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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