Bai 1 : Tinh gia tri cac bieu thuc
a) x2 + 10x + 26 vs x=45
b) x2 - 0,2x + 0,01 vs x= 1,1
Bai 2: Tim x
a) (2x - 5)2 - (4x - 1) . (x + 3) = 5
b) (3x + 4 )2 - (3x - 1) . (3x + 1)= 2
giup mk, mk tick cho !! camon m.n nhieu@@
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
Tim nghien cua da thuc
a)A(x)=-4x-5 h)K(x)=/3x-2/+/4-6x/
b)B(x)=3(2x-1)-2(x + 1) i)M(x)=/x-1/+(x2-1)2
c)C(x)=(2x2-8)(-x2+1) j)N(x)=4x2-3x+7
d)D(x)=3x-x3 k)Pk(x)=7x2-2x-9
l)Q(x)=5x2-11x+6
e)E(x)=2x3+4x
f)G(x)=x3-x2+x-1
a) Đặt A(x)=0
\(\Leftrightarrow-4x-5=0\)
\(\Leftrightarrow-4x=5\)
hay \(x=-\dfrac{5}{4}\)
b) Đặt B(x)=0
\(\Leftrightarrow3\left(2x-1\right)-2\left(x+1\right)=0\)
\(\Leftrightarrow6x-3-2x-2=0\)
\(\Leftrightarrow4x=5\)
hay \(x=\dfrac{5}{4}\)
tim cac gia tri cua x de bieu thuc sau nhan gia tri duong
a) x mu 2 + 4x b)2x mu 2 - 4x c)(x-3)(x+7) d)5(3x +1 )(4x-3) e)(1/2-x)(1/3-x)
cho bieu thuc P= (\(\frac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}+\frac{1}{\sqrt{x}-1}+\frac{1}{\sqrt{x}-3}\) ): \(\frac{1}{x-1}\)
a) Tim dieu kien de P co nghia, rut gon bieu thuc P.
b) Tim cac so tu nhien x de \(\frac{1}{P}\)la so tu nhien
c) Tinh gia tri cua P voi x= 4-\(2\sqrt{3}\)
Giup mk vs mk dang can gap
Tìm x
a) x2(x + 5) - 9x = 45
b) (1 - 2x)2 = (3x - 2)2
\(a,\Leftrightarrow x^2\left(x+5\right)-9\left(x+5\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\\x=-3\end{matrix}\right.\\ b,\Leftrightarrow\left[{}\begin{matrix}1-2x=3x-2\\2x-1=3x-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=1\end{matrix}\right.\)
a) \(\Leftrightarrow x^2\left(x+5\right)-9x-45=0\)
\(\Leftrightarrow x^2\left(x+5\right)-9x\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x^2-9\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-3=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=3\\x=-3\end{matrix}\right.\)
Vậy...
b) \(\Leftrightarrow\left(1-2x\right)^2-\left(3x-2\right)^2=0\)
\(\Leftrightarrow\left(1-2x-3x+2\right)\left(1-2x+3x-2\right)=0\)
\(\Leftrightarrow\left(3-5x\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3-5x=0\\x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=1\end{matrix}\right.\)
Vậy...
Tim Gia tri nho nhat cua ca bieu thuc sau
D=x^2+x+1
E=2x^2-6x
F=3x^2+4x+5
bai 1 tim nguyen de cac bieu thuc sau dat gia tri lon nhat
P=2010-(x+1)^2008
Q=1010-/3x/
C=5 phần (x-3)^2+1
D=4 phần/x-2/+2
dang I. Tim gia tri bieu thuc
Bai 5: Cho x - y =9, tinh gia tri cua bieu thuc
B=\(\frac{4x-9}{3x+y}\)-\(\frac{4y+9}{3y+x}\)(x khac -3y, y khac -3x)
Bai 6: Hay viet cac da thuc duoi dang tong cua cac don thuc roi thu gon. tinh gia tri bieu thuc khi x=-1, y=1
a,D=4x(x+y)-5y(x-y)-4x\(^2\) b,E=(a-1)(x\(^2\)+1)-x(y+1)+(x+y\(^2\)-a+1)
moi nguoi giup minh di hu hu
Giúp mk vs
Giải các phương trình sau
a. 3-2x = 3(x+1) – x – 2
b. (3x+2)(4x-5) = 0
c. (x + 2) (3x + 1) + x2 = 4
a.
\(3-2x=3\left(x+1\right)-x-2\)
\(\Leftrightarrow3-2x=3x+3-x-2\)
\(\Leftrightarrow-2x-3x+x=3-2-3\)
\(\Leftrightarrow-4x=-2\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy.........
b.
\(\left(3x+2\right)\left(4x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\4x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2}{3}\\x=\dfrac{5}{4}\end{matrix}\right.\)
Vậy........
c.
\(\left(x+2\right)\left(3x+1\right)+x^2=4\)
\(\Leftrightarrow\left(x+2\right)\left(3x+1\right)+x^2-4=0\)
\(\Leftrightarrow\left(x+2\right)\left(3x+1\right)+\left(x+2\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(3x+1+x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(4x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{4}\end{matrix}\right.\)
Vậy...........