Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a/ \(x^2=5\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{5}\\x=-\sqrt{5}\end{matrix}\right.\)
vậy .....
b/ \(x^2-9=0\)
\(\Leftrightarrow x^2=9\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2=3^2\\x^2=\left(-3\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
Vậy .......( nhầm cái ngoặc)
c/ \(x^2+1=0\)
\(\Leftrightarrow x^2=-1\)
Mà \(x^2\ge0\Leftrightarrow x\in\varnothing\)
Vậy ....
d/ \(\left(x-1\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^2=3^2\\\left(x-1\right)^2=\left(-3\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=3\\x-1=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)
Vậy ...
e/ \(\left(2x+3\right)^2=25\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x+3\right)^2=5^2\\\left(2x+3\right)^2=\left(-5\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)
Vậy .....
f/ Ta có :
\(x^2=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=1^2\\x^2=\left(-1\right)^2\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Vậy ...
\(x^2=5\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}\\x=-\sqrt{5}\end{matrix}\right.\)
\(\left(x-1\right)^2=9\)
\(\Rightarrow\left[{}\begin{matrix}x-1=3\\x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\)
\(x^2-9=0\Leftrightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
\(\left(2x+3\right)^2=25\)
\(\Rightarrow\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)
\(x^2+1=0\Rightarrow x^2=-1\Rightarrow x\in\varnothing\)
\(x^2=1\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
a) x= 1/8 hoặc -1/8
b) x= 4
c) x= 2
d) Hương ơi Nhi thấy câu này wrong hay sao ấy.
1,
\(\left(2x+1\right)^3=-0,001\\ \left(2x+1\right)^3=\left(-0.1\right)^3\\ \Leftrightarrow2x+1=-0.1\\ 2x=-1.1\\ x=-\dfrac{11}{10}:2\\ x=-\dfrac{11}{20}\\ Vậy...\)
2,
\(\left(2x-3\right)^4=\left(2x-3\right)^6\\ \Leftrightarrow\left(2x-3\right)^6-\left(2x-3\right)^4=0\\ \Leftrightarrow\left(2x-3\right)^4\cdot\left[\left(2x-3\right)^2-1\right]=0\\ \Rightarrow\left\{{}\begin{matrix}\left(2x-3\right)^4=0\\\left(2x-3\right)^2-1=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x-3=0\\\left(2x-3\right)^2=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}2x=3\\2x-3=1\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{2}\\x=2\end{matrix}\right.\\ Vậyx\in\left\{\dfrac{3}{2};2\right\}\)
3, Làm tương tự câu 2
5,
\(9^x:3^x=3\\ \left(9:3\right)^x=3\\ 3^x=3\\ \Rightarrow x=1\\ Vậy...\)
6,
\(3^x+3^{x+3}=756\\ 3^x+3^x\cdot3^3\\ 3^x\cdot\left(1+27\right)=756\\ 3^x\cdot28=756\\ \Leftrightarrow3^x=27\\ 3^x=3^3\\ \Rightarrow x=3\\ vậy...\)
7,
\(5^{x+1}+6\cdot5^{x+1}=875\\ 5^{x+1}\cdot\left(1+6\right)=875\\ 5^{x+1}\cdot7=875\\ \Leftrightarrow5^{x+1}=125\\ \Leftrightarrow5^{x+1}=5^3\Leftrightarrow x+1=3\\ \Rightarrow x=2\\ Vậy...\)
9,
1) \(\frac{25}{12}.x+\frac{11}{15}=\frac{9}{10}\)
=> \(\frac{25}{12}.x=\frac{9}{10}-\frac{11}{15}\)
=> \(\frac{25}{12}.x=\frac{1}{6}\)
=> \(x=\frac{1}{6}:\frac{25}{12}\)
=> \(x=\frac{2}{25}\)
Vậy \(x=\frac{2}{25}\).
3) \(\frac{29}{12}.\left[x\right]-\frac{5}{6}=\frac{3}{8}\)
=> \(\frac{29}{12}.\left[x\right]=\frac{3}{8}+\frac{5}{6}\)
=> \(\frac{29}{12}.x=\frac{29}{24}\)
=> \(x=\frac{29}{24}:\frac{29}{12}\)
=> \(x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\).
4) \(\left[4x+\frac{3}{4}\right]-\frac{5}{4}=2\)
=> \(\left[4x+\frac{3}{4}\right]=2+\frac{5}{4}\)
=> \(4x+\frac{3}{4}=\frac{13}{4}\)
=> \(4x=\frac{13}{4}-\frac{3}{4}\)
=> \(4x=\frac{5}{2}\)
=> \(x=\frac{5}{2}:4\)
=> \(x=\frac{5}{8}\)
Vậy \(x=\frac{5}{8}\).
5) 2x + 2x+3 = 144
⇔ 2x + 2x . 23 = 144
⇔ 2x . (1 + 23) = 144
⇔ 2x . 9 = 144
⇔ 2x = 144 : 9
⇔ 2x = 16
⇔ 2x = 24
=> x = 4
Vậy x = 4.
Chúc bạn học tốt!
\(\left(\frac{3}{5}\right)^{x+1}=\frac{9}{25}\)
\(\left(\frac{3}{5}\right)^{x+1}=\left(\frac{3}{5}\right)^2\)
\(\Rightarrow x+1=2\)
\(\Rightarrow x=1\)
\(\left(2x+1\right)^3=\frac{1}{125}\)
\(\left(2x+1\right)^3=\left(\frac{1}{5}\right)^3\)
\(\Rightarrow2x+1=\frac{1}{5}\)
\(\Rightarrow2x=\frac{-4}{5}\)
\(\Rightarrow x=\frac{-2}{5}\)
vậy \(x=\frac{-2}{5}\)
a) Ta có: \(\left(x+1\right)^2\ge0\forall x\)
\(\Rightarrow A=\left(x+1\right)^2-3\ge-3\)
Dấu " = " xảy ra khi
\(\left(x+1\right)^2=0\)
\(x+1=0\)
\(x=-1\)
Vậy \(x=-1\)khi \(GTNN=-3\)
B:C: tương tự
d) Ta có: \(\left(2x-1\right)^{18}\ge0\forall x\)
\(\left(y+2\right)^2\ge0\forall y\)
\(\Rightarrow D=\left(2x-1\right)^{18}+\left(y+2\right)^2+7\ge7\)
Dấu " = " xảy ra khi \(\hept{\begin{cases}\left(2x-1\right)^{18}=0\\\left(y+2\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}2x-1=0\\y+2=0\end{cases}\Rightarrow}\hept{\begin{cases}2x=1\\y=-2\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{1}{2}\\y=-2\end{cases}}}\)
Vậy \(x=\frac{1}{2};y=-2\)khi \(GTNN=7\)
e) \(\left|-2x+6\right|\ge0\)
\(\Rightarrow E=\left|-2x+6\right|+12\ge12\)
Dấu " = " xảy ra khi \(\left|-2x+6\right|=0\Rightarrow-2x=-6\Rightarrow x=3\)
Vậy x = 3 khi đạt GTNN = 12
F ; G tương tự
hok tốt!!
+) A=(x+1)2 - 3
Vì (x+1)2 \(\ge\)0 nên (x+1)2 - 3 \(\ge\) - 3 .Dấu "=" xảy ra \(\Leftrightarrow\)(x+1)2 = 0 \(\Leftrightarrow\)x = - 1
Vậy min A = - 3 khi x = -1
+) B=(2x-5)20 + 9
Vì (2x-5)20 \(\ge\)0 nên (2x-5)20+9\(\ge\)9.Dấu "=" xảy ra \(\Leftrightarrow\)(2x - 5)20=0 \(\Leftrightarrow\)x=\(\frac{5}{2}\)
Vậy min B=9 khi x=\(\frac{5}{2}\)
Những phần khác cũng làm tương tự :
+) minC= - 5 khi x=\(\frac{4}{3}\)
+) minD= 7 khi x=\(\frac{1}{2}\)và y= - 2
+) minE=12 khi x=3
+) min F = -17 khi x=5
+) min G = -12 khi x= - 4
đơn giản thôi bạn
3 (x + 25)
= 3x + 3*25 (bạn nhân 3 với từng số trong ngoặc) (*là nhân nhé)
= 3x + 75