(x+1)^4=(2x)^4
Tim x, biet.
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\(\Rightarrow\left|3x+1\right|=4+\dfrac{1}{2}=\dfrac{9}{2}\\ \Rightarrow\left[{}\begin{matrix}3x+1=\dfrac{9}{2}\\3x+1=-\dfrac{9}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=\dfrac{7}{2}\\3x=-\dfrac{11}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{6}\\x=-\dfrac{11}{6}\end{matrix}\right.\)
cam on ban
ban co the giup mik bai nay ko
xin hiểu / là phần
a/ 3 và 1/2 - 4/5 - 8 và 1/2 -1/5 -(2021)^0
b 32^5/ 16^5 - (0.2)^100 * (5)^100 -√8181 + /-21/
c 3/2* 16/21 + 3/2 * 5/21 - (- 1/10) ^0 - /-20/
1.
Đặt \(x+y=a\Rightarrow y=a-x\)
\(\Rightarrow x^2+2x\left(a-x\right)-14\left(a-x\right)-10x+3\left(a-x\right)^2+27=0\)
\(\Leftrightarrow2x^2-4\left(a+1\right)x+3a^2-10a+27=0\)
\(\Delta'=4\left(a+1\right)^2-2\left(3a^2-10a+27\right)\ge0\)
\(\Leftrightarrow-a^2+14a-25\ge0\)
\(\Rightarrow7-2\sqrt{6}\le a\le7+2\sqrt{6}\)
\(\Rightarrow-10-2\sqrt{6}\le P\le-10+2\sqrt{6}\)
2. Chắc đề là \(a;b>0\) (đảm bảo mẫu dương) chứ ko phải \(a.b>4\)
\(M\ge\dfrac{\left(a+b\right)^2}{a+b-8}=\dfrac{\left(a+b-8+8\right)^2}{a+b-8}=\dfrac{\left(a+b-8\right)^2+16\left(a+b-8\right)+64}{a+b-8}\)
\(M\ge a+b-8+\dfrac{64}{a+b-8}+16\ge2\sqrt{\dfrac{64\left(a+b-8\right)}{a+b-8}}+16=32\)
Dấu "=" xảy ra khi \(a=b=8\)
\(\left|x-\dfrac{1}{2}\right|\left|2x-\dfrac{3}{4}\right|=2x-\dfrac{3}{4}\)
\(\left\{{}\begin{matrix}\left|x-\dfrac{1}{2}\right|\ge0\\\left|2x-\dfrac{3}{4}\right|\ge0\end{matrix}\right.\)
\(\Rightarrow2x-\dfrac{3}{4}\ge0\)
\(\Rightarrow\left|2x-\dfrac{3}{4}\right|=2x-\dfrac{3}{4}\)
\(\Rightarrow\left|x-\dfrac{1}{2}\right|=1\)
\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=1\Rightarrow x=\dfrac{3}{2}\\x-\dfrac{1}{2}=-1\Rightarrow x=-\dfrac{1}{2}\end{matrix}\right.\)
\(2x-\dfrac{3}{4}\ge0\Rightarrow2x\ge\dfrac{3}{4}\Rightarrow x\ge\dfrac{3}{2}\)
Vậy xảy ra khi:
\(x=\dfrac{3}{2}\)
12x/6 + 4x/6 - 3x/6 = 3/8 - 1/4 - 1/6
13x/6 = (9 - 6 - 4)/24 = -1/24
--> x = -1/52
câu trên mk làm rồi
\(\dfrac{2x-1}{x-3}=\dfrac{2x+3}{x-1}\)
\(\Rightarrow\left(2x-1\right)\left(x-1\right)=\left(x-3\right)\left(2x+3\right)\)
\(\Rightarrow2x^2-x-2x+1=2x^2-6x+3x-9\)
\(\Rightarrow-x-2x+6x-3x=-1-9\)
\(\Rightarrow0=-10\) (vô lí)
Vậy ko tồn tại giá trị của x.
\(\left(x+1\right)^4=\left(2x\right)^4\)
\(\Rightarrow x+1=2x\)
\(1=2x-x\)
\(\Rightarrow x=1\)
\(\left(x+1\right)^4=\left(2x\right)^4\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=2x\\x+1=-2x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-2x=-1\\x+2x=-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-x=-1\\3x=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{1}{3}\end{cases}}\)