\(\left(\frac{2}{5}-3x\right)^2=\frac{9}{25}\)
tìm x nha mọi người
giải hộ mình với
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(2x+\frac{3}{5}\right)^2-\frac{9}{25}=0\)
\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\frac{9}{25}\)
\(\Leftrightarrow\left(2x+\frac{3}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}2x+\frac{3}{5}=\frac{3}{5}\\2x+\frac{3}{5}=-\frac{3}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=0\\2x=-\frac{6}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{3}{5}\end{cases}}\)
_Tần vũ_
\(3\left(3x-\frac{1}{2}\right)^3+\frac{1}{9}=0\)
\(\Leftrightarrow3\left(3x-\frac{1}{2}\right)^3=-\frac{1}{9}\)
\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=-\frac{1}{27}\)
\(\Leftrightarrow\left(3x-\frac{1}{2}\right)^3=\left(-\frac{1}{3}\right)^3\)
\(\Leftrightarrow3x-\frac{1}{2}=\frac{-1}{3}\)
\(\Leftrightarrow3x=\frac{1}{6}\)
\(\Leftrightarrow x=\frac{1}{18}\)
_Tần Vũ_
\(VT=\left|3x+1\right|+\left|3x-5\right|=\left|3x+1\right|+\left|5-3x\right|\ge\left|3x+1+5-3x\right|=6\)
\(VP=\frac{12}{\left(y+3\right)^2+2}\le\frac{12}{2}=6\)
Như vậy \(VT\ge6;VP\le6\)
Mà \(VT=VP\Leftrightarrow VT=VP=6\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}-\frac{1}{3}\le x\le\frac{5}{3}\\y=-3\end{cases}}\)
\(11-\left(3x-1\right)=\frac{9}{2}-\left(5-3,5x\right)\)
\(=>11-3x+1=\frac{9}{2}-5+3,5x\)
\(=>-3x+12=3,5x-\frac{1}{2}\)
\(=>-3x-3,5x=-\frac{1}{2}-12\)
\(=>-6,5x=-12,5\)
\(=>x=\frac{-12,5}{-6,5}=\frac{25}{13}\)
Ủng hộ nha
\(11-\left(3x-1\right)=\frac{9}{2}-\left(5-3,5x\right)\)
\(11-3x+1=\frac{9}{2}-5+3,5x\)
\(12-3x=-\left(0,5\right)+3,5x\)
\(12,5-3x=3,5x\)
\(12,5=6,5x\)
\(x=12,5:6,5=\frac{25}{13}\)
5/9 : (1/11 - 5/12) + 5/9 : (1/15 - 2/3)
= 5/9 : (-43/132) + 5/9 : (-3/5)
= 5/9 . (-132/43) + 5/9 . (-5/3)
= 5/9 . [-132/43 + (-5/3)]
= 5/9 . 445/129
= 2225/1161.
Ai k mình mình k lại.
a,<=>\(\frac{\left(2x+1\right)^2}{4}\)+\(\frac{2\left(2x-1\right)^2}{4}\)≥\(\frac{12\left(x+5\right)^2}{4}\)
<=>4x2+4x+1+2(4x2-4x+1)≥12(x2+10x+25)
<=>4x2+4x+1+8x2-8x+2≥12x2+120x+300
<=>4x2+4x+1+8x2-8x+2-12x2-120x-300≥0
<=>-124x-297≥0
<=>124x+297≤0
<=>124x≤-297
<=>x≤\(\frac{-297}{124}\)
b, Tương tự câu a
c, |5−3x|=2+x
TH1: 5-3x=2+x
<=> -3x - x = 2 - 5
<=> -4x = -3
<=> x = 3/4
TH2: 5-3x = -2 - x
<=> -3x + x = -2 - 5
<=> -2x = -7
<=> x = 7/2
a,
\(\left(\frac{1}{2}\right)^{2x+1}=\frac{1}{32}\)
\(\left(\frac{1}{2}\right)^{2x+1}=\left(\frac{1}{2}\right)^5\)
=>\(2x+1=5\)
2x=5-1
2x=4
x=4:2
x=2
b, mình không biết cách làm
a)\(\left(\frac{1}{2}\right)^{2x+1}=\frac{1}{32}\)
\(\left(\frac{1}{2}\right)^{2x+1}=\left(\frac{1}{2}\right)^5\)
\(\Rightarrow2x+1=5\)
\(\Rightarrow x=2\)