cac ban giup mik nha
1) 10x2(x-1)+15xy(1-x)
2) 25a2-x2+10x-2
LH
Những câu hỏi liên quan
giup mik voi cac ban
25a2-25x2+10x-1
\(25a^2-25x^2+10x-1\)
\(=25a^2-\left(5x-1\right)^2\)
\(=\left(5a-5x+1\right)\left(5a+5x-1\right)\)
Đúng 1
Bình luận (1)
\(25a^2-25x^2+10x-1=25a^2-\left(25x^2-10x+1\right)=\left(5a\right)^2-\left(5x-1\right)^2=\left(5a+5x-1\right)\left(5a-5x+1\right)\)
Đúng 1
Bình luận (1)
Cho $x^2-3x+1=0$x2−3x+1=0
Tính $A=\frac{\left(x^4+x^3-10x^2+x+2015\right)\left(x^4+x^2+1\right)+x^4+3x^2+1}{x^4+x^2+1}$A=(x4+x3−10x2+x+2015)(x4+x2+1)+x4+3x2+1x4+x2+1
Các bn giải giúp mik với, ko pik đề có bị sai ko mà lm ko đc
a : (x-\(\dfrac{1}{2}\))^2=0
b: (x-2)^2=1
c: (2x-1)^3=-8
d: (x+\(\dfrac{1}{2}\))^2=\(\dfrac{1}{16}\)
cac ban giup mik nha mik ko biet cach trinh bay
giup mik mik dang can gap
caam on cac ban nhieu
a) \(\left(x-\dfrac{1}{2}\right)^2=0\)
\(\Rightarrow x-\dfrac{1}{2}=0\)
\(\Rightarrow x=\dfrac{1}{2}\)
b) \(\left(x-2\right)^2=1\)
\(\Rightarrow x-2=1\)
\(\Rightarrow x=3\)
c) \(\left(2x-1\right)^3=-8\)
\(\Rightarrow\left(2x-1\right)^3=\left(-2\right)^3\)
\(\Rightarrow2x-1=-2\)
\(\Rightarrow2x=-1\)
\(\Rightarrow x=\dfrac{-1}{2}\)
d) \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)
\(\Rightarrow\left(x+\dfrac{1}{2}\right)^2=\left(\dfrac{1}{4}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{4}\\x+\dfrac{1}{2}=-\dfrac{1}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{4}\\x=\dfrac{-3}{4}\end{matrix}\right.\).
Đúng 0
Bình luận (3)
a , \(\left(x-\dfrac{1}{2}\right)^2=0\)
<=> \(x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\)
b , \(\left(x-2\right)^2=1\Rightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
c , \(\left(2x-1\right)^3=-8\Rightarrow2x-1=-2\Rightarrow x=\dfrac{-1}{2}\)
d , \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{4^2}\)
<=> \(\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{4}\\x+\dfrac{1}{2}=\dfrac{-1}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{4}\\x=\dfrac{-3}{4}\end{matrix}\right.\)
Đúng 0
Bình luận (1)
a) \(\left(x-\dfrac{1}{2}\right)^2=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0^2\)
\(\Leftrightarrow x-\dfrac{1}{2}=0\)
\(\Leftrightarrow x=0+\dfrac{1}{2}\)
\(\Leftrightarrow x=\dfrac{1}{2}\left(TM\right)\)
Vậy \(x=\dfrac{1}{2}\) là giá trị cần tìm
b) \(\left(x-2\right)^2=1\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-2\right)^2=1^2\\\left(x-2\right)^2=\left(-1\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=1\\x-2=\left(-1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=1\end{matrix}\right.\) \(\left(TM\right)\)
Vậy \(x\in\left\{3;1\right\}\)
c) \(\left(2x-1\right)^3=-8\)
\(\Rightarrow\left(2x-1\right)^3=\left(-2\right)^3\)
\(\Rightarrow2x-1=-2\)
\(\Rightarrow2x=-2+1\)
\(\Rightarrow2x=-1\)
\(\Rightarrow x=\dfrac{-1}{2}\left(TM\right)\)
Vậy \(x=\dfrac{-1}{2}\)
d) \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{4}\\\left(x+\dfrac{1}{2}\right)^2=\dfrac{-1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{4}\\x+\dfrac{1}{2}=-\dfrac{1}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-1}{4}\\x=\dfrac{-3}{4}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{-1}{4};\dfrac{-3}{4}\right\}\) là giá trị cần tìm
Đúng 0
Bình luận (4)
Xem thêm câu trả lời
a)tim x
1/(3x-1)2-(x+7)2=0
giup mik nha cac ban
Ta có : \(\left(3x-1\right)^2-\left(x+7\right)^2=0\)
\(\left(3x-1+x+7\right)\left(3x-1-x-7\right)=0\)
\(\left(4x+6\right)\left(2x-8\right)=0\)
Nên : 4x + 6 = 0 hoặc 2x - 8 = 0
4x = -6 hoặc 2x = 8
x = \(\frac{-3}{2}\) hoặc x = 4
Vậy x = \(\frac{-3}{2}\) hoặc x = 4
Tim cac so nguyen x,y biet:a)*x2 -3*x+1 chia het cho x+2 b)x2-xy = 5x-4y-9 c) (x2-8)*(x2-15)<0 d) (x+1)2+(y+1)2+(x-y)2=2 e) (x2-4)x2>0 ai lam truoc minh tick,mai minh di hoc rui,lam giup minh di pls Thank ban truoc ne ;(
bài 1 Tìm GTLN của biểu thức
A = 11-10x-x2
minh xin loi nhe. cac ban giup minh
A=-(11+5x+5x+x^2)
A=-((5x+x^2)+(5x+20)-9)
A=-(x(5+x)+5(5+x)-9)
A=-((5+x)^2-9)
A=(5+x)^2+9
Max A= 9 Khi x=-5
Đúng 0
Bình luận (0)
S = 1 x 2+2x3+3x4+...39x40
S=...
Giup mik nha cac ban
3S=1x2x3+2x3x3+...+39x40x3
3S=1x2x(3-0)+2x3x(4-1)+...+39x40x(41-38)
3S=1x2x4-1x2x0+2x3x4-2x3x1+.....+39x40x41-38x39x40
3S=39x40x41
S=39x40x41:3
S=13x40x41
S=21320
Đúng 0
Bình luận (0)
S = 1 x 2 + 2 x 3 + 3 x 4 + ... + 39 x 40
S = 2 + 6 + 12 + ... + 1560
Sau đó tự làm nhé !
Đúng 0
Bình luận (0)
bài 1 Tìm GTNN của biểu thức
A= 25x2+3y2-10x+11
B= (x-3)2+(x-11)2
C= (x+1).(x-2).(x-3).(x-6)
cac ban giup minh nhe. minh dang can
C=[(x+1)(x-6)][(x-2)(x-3)]
=(x2-5x-6)(x2-5x+6)
=(x2-5x)2-36>=-36
GTNN cua C=-36 tai x2-5x=0=>x(x-5)=0=>x=0 hoac x=5
Đúng 0
Bình luận (0)
B=(x-3)2+(x-11)2
=x2-6x+9+x2-22x+121
=2x2-28x+130
=2(x2-14x+65)
=2(x2-2.7x+72-72+65)
=2[(x-7)2-49+65]
=2(x-7)2+32
=> vì 2(x-7)2 >= 0
=>2(x-7)2+32 >= 32
=> GTNN của B=32. Khi x=7
Đúng 0
Bình luận (0)
A=25x2+3y2-10x+11
=25x2-10x+1+3y2+10
=(5x-1)2+3y2+10
=>vì (5x-1)2+3y2 >= 0
=>(5x-1)2+3y2+10>= 10
=> GTNN của A=10 . Khi x=1/5 và y=0
Đúng 0
Bình luận (0)
Xem thêm câu trả lời
Cac ban giup minh voi
1) Giai cac phuong trinh
a) 2010.(4x-3)-4x2+3=0
b)( x2-\(\frac{25}{4}\))2= 10x +1