Tim x
-45: (3x - 17) = 3 mu 2
Help me
tìm cực trị:y=(1-x)^3(3x-8)^2
help me!!!
(x-1)^2 = (2x-3)^2
help me
\(\Rightarrow\left(x-1\right)^2-\left(2x-3\right)^2=0\\ \Rightarrow\left(x-1-2x+3\right)\left(x-1+2x-3\right)=0\\ \Rightarrow\left(2-x\right)\left(3x-4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{4}{3}\end{matrix}\right.\)
1) x^2-3x-1
2) 3x^2-5x-2
help
1:Ta có: \(x^2-3x-1\)
\(=x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{13}{4}\)
\(=\left(x-\dfrac{3}{2}\right)^2-\dfrac{13}{4}\ge-\dfrac{13}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{3}{2}\)
tim x
3x - 4 mu 5 : 4 mu 3= 2
\(3x-4^5:4^3=2\\ \Rightarrow3x-4^2=2\\ \Rightarrow3x=18\\ \Rightarrow x=6\)
3x-4^5:4^3=2
3x-4^2=2
3x-16 =2
3x =16+2
3x=18
x =18:3
x=6
tim STN x biet
a, 21 thuoc B(x-3)
b, 1-x thuoc U(17)
c, 2x+3 thuoc B(2x-1)
d, x+1 thuoc U(x mu 2+x+3)
e, 3x+1:11-2x
a: \(\Leftrightarrow x-3\inƯ\left(21\right)\)
\(\Leftrightarrow x-3\in\left\{-3;-1;1;3;7;21\right\}\)
hay \(x\in\left\{0;2;4;6;10;24\right\}\)
b: \(\Leftrightarrow x-1\in\left\{1;-1;17\right\}\)
hay \(x\in\left\{2;0;18\right\}\)
c: \(\Leftrightarrow2x-1+4⋮2x-1\)
\(\Leftrightarrow2x-1\in\left\{1;-1\right\}\)
hay \(x\in\left\{1;0\right\}\)
d: \(\Leftrightarrow x^2+x+3⋮x+1\)
\(\Leftrightarrow x+1\inƯ\left(3\right)\)
\(\Leftrightarrow x+1\in\left\{1;3\right\}\)
hay \(x\in\left\{0;2\right\}\)
x^2+1/x+x/x^2+1=-5/2
help me! thanks
tim so tu nhien x , biet
a, 5x - x = 29 - 36 : 4
b, ( 3x - 48 ) . 6 = 3 mu 3 . 2 mu 2 + 2 mu 3 . 3 mu 2 - 12 mu 2
c, x thuoc B(6) va 22 < x < hoac = 36
a) \(\Leftrightarrow4x=20\Leftrightarrow x=5\)
b) \(\Leftrightarrow18x-288=27.4+8.9-144\)
\(\Leftrightarrow18x=108+72-144+288=324\)
\(\Leftrightarrow x=18\)
c)Ta có:
\(x\in B\left(6\right)\) và \(22< x\le36\)
mà \(B\left(6\right)=\left\{0;6;12;18;24;30;36;42;48;...\right\}\)
\(\Rightarrow x\in\left\{0;6;12;18;24;30;36;42;48;...\right\}\) và \(22< x\le36\)
\(\Rightarrow x\in\left\{24;30;36\right\}\)
Vậy \(x\in\left\{24;30;36\right\}\)
tim x nguyen biet:
a 8.(x mu 2 +3).(5-x)
b)(2x + 1)mu 2=25
c) (1-3x)mu3 =64
d)(4-x)mu3 =-27
e) xmu2 -5x =0
b: \(\left(2x+1\right)^2=25\)
=>\(\left[{}\begin{matrix}2x+1=5\\2x+1=-5\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}2x=4\\2x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
c: \(\left(1-3x\right)^3=64\)
=>\(\left(1-3x\right)^3=4^3\)
=>1-3x=4
=>3x=1-4=-3
=>x=-3/3=-1
d: \(\left(4-x\right)^3=-27\)
=>\(\left(4-x\right)^3=\left(-3\right)^3\)
=>4-x=-3
=>x=4+3=7
e: \(x^2-5x=0\)
=>\(x\left(x-5\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
a/[x-2][7-x] > 0 . tim x thuoc Z
b/[x mu 2+13][x mu 2-17] > 0 . tim x thuoc Z
c/[x mu 2-13][x mu 2-17] <0 . tim x thuoc Z
GIUP MINH VOI
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CAM ON CAC BAN NEU CAC BAB GIUP MINH
a, => [x-2] và [7-x] cùng dấu
Xét 2 trường hợp cùng >0 và cùng<0
b, tương tự
c, xét 2 trường hợp khác dấu
Có gì ko h bạn cứ hỏi nha!