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a: \(\Leftrightarrow2^x=1024\cdot3+1024\cdot7776+7776\cdot5\)
\(\Leftrightarrow2^x=8004576\)
hay \(x\in\varnothing\)
b: \(\Leftrightarrow x\left(x+3\right)^{100}-\left(x+3\right)^{100}=0\)
\(\Leftrightarrow\left(x+3\right)^{100}\left(x-1\right)=0\)
=>x=-3 hoặc x=1
1.Tim x:
a)| x + 1 | = 5 -> Th1: x+1=5-> x= 5-1=4
Th2: x+1=-5-> x= (-5) -1=-6(Loại. vì x lớn hơn hoặc bằng 0)
Vậy x= 4
b)| x - 3 | = 7 -> TH1: x-3=7-> x=7+3=10(Loại. Vì x<3)
TH2: x-3=-7-> x=-7+3=-4
Vậy x= -4
c) x + | 2 - x | = 6
-> | 2 - x | =6 -x
-> TH1: 2-x = 6-x
-> -x+ x= 2-6
-> 0x =-4(LOẠI)
TH2: 2-x= -6+x
->(-x)-x= 2+6
-> -2.x=8
-> x=8: -2=-4
Vậy x=-4
Tick cho mik nha!!!
2. Tìm x
a) | x | = 7-> x=-7 hoặc x=7
b) | x | < 7.Vì| x | lớn hơn hoặc bằng 0
-> | x | =(0;1;2;3;4;5;6)
-> x= (-6;-5;-4;-3;-2;-1;0;1;2;3;4;5;6)
c) | x | > 7
-> | x | =(8;9;10;11;12;13.............)
-> x= (...............;-9;-8;8;9;10;.............)
Bài 1 tự làm!
Bài 2:
a, \(\left(3x-4\right)\left(x-1\right)^3=0\Rightarrow\left[{}\begin{matrix}3x-4=0\\\left(x-1\right)^3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=1\end{matrix}\right.\)
b, \(2^{2x-1}:4=8^3\Rightarrow2^{2x-1}:2^2=2^9\)
\(\Rightarrow2x-1-2=9\Rightarrow2x-3=9\Rightarrow2x-12\Rightarrow x=6\)
c, Đề chưa rõ
d, \(\left(x+2\right)^5=2^{10}\Rightarrow\left(x+2\right)^5=4^5\Rightarrow x+2=4\Rightarrow x=2\)
e, \(\left(3x-2^4\right).7^3=2.7^4\Rightarrow3x-2^4=2.7^4:7^3\Rightarrow3x-16=2.7=14\)
\(\Rightarrow3x=14+16=30\Rightarrow x=\dfrac{30}{3}=10\)
f, \(\left(x+1\right)^2=\left(x+1\right)^0\Rightarrow\left(x+1\right)^2=1\) (vì x0 = 1)
\(\Rightarrow x+1=1\Rightarrow x=0\)
a) \(\left(3x-\frac{1}{2}\right)^2=\frac{1}{121}=\left(\frac{1}{11}\right)^2\)
=> \(\orbr{\begin{cases}3x-\frac{1}{2}=\frac{1}{11}\\3x-\frac{1}{2}=-\frac{1}{11}\end{cases}}\)
=> \(\orbr{\begin{cases}3x=\frac{13}{22}\\3x=\frac{9}{22}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{13}{66}\\x=\frac{3}{22}\end{cases}}\)
b) \(\left(5-3x\right)^3=\left(-\frac{1}{27}\right)=\left(-\frac{1}{3}\right)^3\)
=> \(5-3x=-\frac{1}{3}\)
=> \(3x=\frac{16}{3}\)
=> \(x=\frac{16}{3}:3=\frac{16}{9}\)
c) 5x + 5x+2 = 650
=> 5x + 5x . 52 = 650
=> 5x(1 + 52) = 650
=> 5x . 26 = 650
=> 5x = 25
=> 5x = 52 => x = 2
d) 3x-1 + 5.3x-1 = 126
=> (1 + 5).3x-1 = 126
=> 6.3x-1 = 126
=> 3x-1 = 21
=> 3x-1 =3.7
tới đây là không xử lí được x luôn :)
a,\(\left(3x-\frac{1}{2}\right)^2=\frac{1}{121}=\left(\frac{1}{11}\right)^2=\left(-\frac{1}{11}\right)^2\)
\(< =>\orbr{\begin{cases}3x-\frac{1}{2}=\frac{1}{11}\\3x-\frac{1}{2}=-\frac{1}{11}\end{cases}}< =>\orbr{\begin{cases}3x=\frac{1}{11}+\frac{1}{2}\\3x=-\frac{1}{11}+\frac{1}{2}\end{cases}}\)
\(< =>\orbr{\begin{cases}3x=\frac{2}{22}+\frac{11}{22}=\frac{13}{22}\\3x=\frac{11}{22}-\frac{2}{22}=\frac{9}{22}\end{cases}}\)
\(< =>\orbr{\begin{cases}x=\frac{13}{22}:3=\frac{13}{22}.\frac{1}{3}=\frac{13}{66}\\x=\frac{9}{22}:3=\frac{9}{22}.\frac{1}{3}=\frac{9}{66}=\frac{3}{22}\end{cases}}\)
b,\(\left(5-3x\right)^2=-\frac{1}{27}=\left(-\frac{1}{3}\right)^3\)
\(< =>5-3x=-\frac{1}{3}< =>-3x=-\frac{1}{3}-5=-\frac{16}{3}\)
\(< =>3x=\frac{16}{3}< =>x=\frac{16}{3}:3=\frac{16}{3}.\frac{1}{3}=\frac{16}{9}\)
c,\(5^x+5^{x+2}=650< =>5^x+5^x.25=650\)
\(< =>5^x\left(25+1\right)=5^x=\frac{650}{36}=25< =>x=2\)
bạn nào giúp câu d
Ta có : \(\left(x-5\right)^4=\left(x-5\right)^6\)
\(\Rightarrow\left(x-5\right)^4-\left(x-5\right)^6=0\)
\(\Leftrightarrow\left(x-5\right)^4\left[1-\left(x-5\right)^2\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x-5\right)^4=0\\1-\left(x-5\right)^2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\\left(x-5\right)^2=1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\\\left(x-5\right)=-1;1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=4;6\end{cases}}\)
Vậy x = {4;5;6}.
b) Ta có : ax = a50
=> x = 50
d) Ta có : 1 + 2 + 3 + ..... + x = 222111
=> \(\frac{\left[\left(x-1\right):1+1\right]\left(x+1\right)}{2}=222111\)
=> \(\frac{x\left(x+1\right)}{2}=222111\)
=> x(x + 1) = 444222
=> x(x + 1) = 666.667
=> x = 666
Vậy x = 666.