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\(\left(5-x\right)^3=-64\)
\(\Leftrightarrow\left(5-x\right)^3=\left(-4\right)^3\)
\(\Leftrightarrow5-x=-4\)
\(\Leftrightarrow x=9\)
1)
( 5 - x )3 = - 64
( 5 - x )3 = ( - 4 )3
\(\Rightarrow\)5 - x = - 4
x = 5 - ( - 4 )
x = 5 + 4
x = 9
2)
A = 4 + 42 +43 + 44 + ........ + 499 + 4100
A = 4(4+1) + 43(4+1) + 45(4+1) + ........... + 499(4+1)
A = 4 . 5 + 43 . 5 + 45 . 5 + ............ + 499 . 5
A = 5(4+43 + 45 + ..........+ 499 )
Mà 5 \(⋮\) 5 \(\Rightarrow\) 5(4+43 + 45 + ........ + 499 ) \(⋮\) 5
Vậy A chia 5 dư 0
a) \(326-2x=78\Rightarrow2x=326-78=248\)
\(\Rightarrow x=248:2=124\)
b) \(42.5-\left(3x+6\right)=3^4\Rightarrow210-\left(3x+6\right)=81\)
\(\Rightarrow3x+6=210-81=129\Rightarrow3x=129-6=123\)
\(\Rightarrow x=123:3=41\)
a, 326 -2x =78
\(\Rightarrow\)2x=326-78=248
\(\Rightarrow\)x=248/2=124
b,\(42.5-\left(3x+6\right)=3^4\)
\(\Rightarrow\)\(210-3x-6=81\)
\(\Rightarrow\)3x=123
\(\Rightarrow\)x=123/3=41
c, \(4^x:64=4^{250}\)
\(\Rightarrow\)\(4^x:4^3=4^{250}\)
\(\Rightarrow\)x=253
d, \(30x+1+2+3+...+30=495\)
\(\Rightarrow\)\(30x+465=495\)
\(\Rightarrow\)\(x=1\)
a: \(\Leftrightarrow x^3=\dfrac{539}{64}\)
hay \(x=\dfrac{7\sqrt{11}}{4}\)
c: \(\Leftrightarrow2^{2x-1}=2^9\cdot2^2=2^{11}\)
=>2x-1=11
hay x=6
d: \(\Leftrightarrow x^{17}-x=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)
hay \(x\in\left\{0;1;-1\right\}\)
a)\(3^2.3^x=243\)
\(3^2.3^x=3^5\)
\(3^x=3^5:3^2\)
\(3^x=3^3\)
\(\Rightarrow x=3\)
b)\(2^x.16^2=1024\)
\(2^x.\left(2^4\right)^2=2^{10}\)
\(2^x.2^8=2^{10}\)
\(2^x=2^{10}:2^8\)
\(2^x=2^2\)
\(\Rightarrow x=2\)
Học tốt nha^^
3^2.3=243 vậy x=3
2^x.16^2=1024 vậy x=2
64.4^x=16^8 vậy x=13
2^x-26=6 vậy x=5
phần cuối đề sai
a) \(\left|x\right|-\frac{7}{6}=\frac{9}{15}\)
=> \(\left|x\right|=\frac{9}{15}+\frac{7}{6}=\frac{53}{30}\)
=> \(\orbr{\begin{cases}x=\frac{53}{30}\\x=-\frac{53}{30}\end{cases}}\)
b) \(\left|x-\frac{4}{3}\right|=\frac{1}{6}\)
=> \(\orbr{\begin{cases}x-\frac{4}{3}=\frac{1}{6}\\x-\frac{4}{3}=-\frac{1}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{7}{6}\end{cases}}\)
c) \(\left|x-\frac{4}{3}\right|-\frac{1}{3}=\frac{1}{2}\)
=> \(\left|x-\frac{4}{3}\right|=\frac{1}{2}+\frac{1}{3}\)
=> \(\left|x-\frac{4}{3}\right|=\frac{5}{6}\)
=> \(\orbr{\begin{cases}x-\frac{4}{3}=\frac{5}{6}\\x-\frac{4}{3}=-\frac{5}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{13}{6}\\x=\frac{1}{2}\end{cases}}\)
d) \(\frac{8}{3}-\left|\frac{7}{9}-x\right|=-\frac{1}{5}\)
=> \(\left|\frac{7}{9}-x\right|=\frac{43}{15}\)
=> \(\orbr{\begin{cases}\frac{7}{9}-x=\frac{43}{15}\\\frac{7}{9}-x=-\frac{43}{15}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{94}{45}\\x=\frac{164}{45}\end{cases}}\)
e) \(\left|x-\left(\frac{1}{4}\right)^2\right|-\frac{25}{64}=0\)
=> \(\left|x-\frac{1}{16}\right|=\frac{25}{64}\)
=> \(\orbr{\begin{cases}x-\frac{1}{16}=\frac{25}{64}\\x-\frac{1}{16}=-\frac{25}{64}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{29}{64}\\x=-\frac{21}{64}\end{cases}}\)
f) \(\left(x-\frac{1}{4}\right)^2+\frac{17}{64}=\frac{21}{32}\)
=> \(\left(x-\frac{1}{4}\right)^2=\frac{25}{64}\)
=> \(\left(x-\frac{1}{4}\right)^2=\left(\frac{5}{8}\right)^2\)
=> \(\orbr{\begin{cases}x-\frac{1}{4}=\frac{5}{8}\\x-\frac{1}{4}=-\frac{5}{8}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{7}{8}\\x=-\frac{3}{8}\end{cases}}\)
\(\left(\frac{2^3}{4}\right).2^{x+1}=64\)
\(\left(\frac{2^3}{2^2}\right).2^{x+1}=64\)
\(2.2^{x+1}=64\)
\(2^{x+1+1}=2^6\)
\(\Rightarrow x+2=6\)
\(x=6-2=4\)
( 23 : 4 ) . 2( x + 1 ) = 64
( 8 : 4 ) . 2( x + 1 ) = 64
2 . 2( x + 1 ) = 64
2( x + 1 ) = 64 : 2
2( x + 1 ) = 32
ta thấy 32 = 8 .4 = 2.2.2.2.2= 25
=> 2( x + 1 ) = 25
=> x + 1 = 5
x = 5 - 1 = 4
Vậy x = 4