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a) \(9.x-2.x=\frac{6^{27}}{6^{25}}+\frac{48}{12}\)
\(\Leftrightarrow7x=6^2+4\)
\(\Leftrightarrow7x=36+4=40\)
\(\Leftrightarrow x=\frac{40}{7}\)
Vậy : \(x=\frac{40}{7}\)
b) \(11^x=5.x+\frac{5^{31}}{5^{29}}+3.2^2-10^0\)
\(\Leftrightarrow11^x=5x+5^2+12-1\)
\(\Leftrightarrow11^x=5x+36\)
\(\Rightarrow x\in\varnothing\)
=9x2^25-2^2x2^26/2^24x5^2-2^27x3
=9x2^25-2^28/2^24x5^2-2^27x3
=2^25x(9-2^3)/2^24x(5^2-2^3x3)
=2^25/2^24
=2^1=2
\(a,\left(-5\right).\left|x\right|=-75\)
\(\left|x\right|=\frac{-75}{-5}=15\)
\(\Rightarrow\orbr{\begin{cases}x=15\\x=-15\end{cases}}\)
Vậy....
\(b,\left(-6\right)^3.x^2=-1944\)
\(-216.x^2=-1944\)
\(x^2=9\)
\(\Rightarrow x=\pm3\)
Vậy....
\(d,\left|9-x\right|=-7+64\)
\(\left|9-x\right|=57\)
\(\Rightarrow\orbr{\begin{cases}9-x=57\\9-x=-57\end{cases}\Rightarrow\orbr{\begin{cases}x=-48\\x=66\end{cases}}}\)
Vậy...
\(e,\left|x+101\right|-\left(-16\right)=\left(-43\right).\left(-5\right)\)
\(\left|x+101\right|+16=215\)
\(\left|x+101\right|=199\)
\(\Rightarrow\orbr{\begin{cases}x+101=199\\x+101=-199\end{cases}\Rightarrow\orbr{\begin{cases}x=98\\x=-300\end{cases}}}\)
Vậy..
hok tốt!!
a,\(\left(-5\right).\left|x\right|=-75\)
\(=>\left|x\right|=-75:\left(-5\right)=15\)
\(=>\orbr{\begin{cases}x=15\\x=-15\end{cases}}\)
b,\(\left(-6\right)^3.x^2=-1944\)
\(=>\frac{1944}{216}=x^2\)
\(=>x=\sqrt{\frac{1944}{216}}=3\)
Mik chỉ làm 1 câu chung cho bài 1 thôi nha , mấy câu sau giống .
Tìm x , biết :
a) ( x + 1) 2 . ( x - 2 )2 = 0
=> \(\left\{{}\begin{matrix}\left(x+1\right)^2=0\\\left(x-2\right)^2=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
Vậy x = -1 hoặc x = 2 .
Bài 2 , rút gọn biểu thức :
A = a.(b -c) - b.(a+c)
= ab - ac - ( ab + bc )
= ab - ac - ab - bc
= ac - bc
= c .(a-b)
C = (a+3b).c - d - (3a-d).(b+c) - 2c.(b - a) + 2b.(a+d)
= ac + 3bc - d - (3a - d).(b+c) - 2cb - 2ca + 2ba + 2bd
= ac + ( 3bc - 2bc ) - d - ( 3a - d) . ( b+c) +(-2ca + 2ba ) +2db
= ac + bc - d - ( 3a -d) . ( b+c) -2a + cb + 2db
= (a+b).c - d - (3a-d) . ( b+c) - 2a + (2d+c).b
= .........(mik chịu )..........
Bài 1:
a) 02002 < 02023
b) 20220 = 20230
c) 549 < 5510
d) ( 4 + 5 )3 > 42 + 52
đ) 92 - 32 > ( 9 - 3 )2
Bài 2:
a) 32 x 43 - 32 + 333
= 9 x 64 - 9 + 333
= 576 - 9 + 333
= 567 + 333
= 900
b) 5 x 43 + 24 x 5 + 410
= 5 x 64 + 24 x 5 + 1
= 5 x ( 64 + 24 ) + 1
= 5 x 88 + 1
= 440 + 1
= 441
c) 23 x 42 + 32 x 5 - 40 x 12023
= 8 x 16 + 9 x 5 - 40 x 1
= 128 + 45 - 40
= 133
Bài 1 :
a) \(0^{2002}=0;0^{2023}=0\Rightarrow0^{2002}=0^{2023}\)
b) \(2022^0=1;2023^0=1\Rightarrow2022^0=2023^0\)
c) \(54^9< 55^9;55^9< 55^{10}\Rightarrow54^9< 55^{10}\)
d) \(\left(4+5\right)^3>\left(4+5\right)^2;\left(4+5\right)^2>4^2+5^2\Rightarrow\left(4+5\right)^3>4^2+5^2\)
đ) \(9^2-3^2=81-9=82;\left(9-3\right)^2=6^2=36\Rightarrow9^2-3^2>\left(9-3\right)^2\)
a)\(51-\left(3+x\right)=26\\ \Leftrightarrow51-3-x=26\\ \Leftrightarrow x=51-3-26\\ \Leftrightarrow x=22\)
b)Ta có:\(Ư_{\left(24\right)}=\left\{1;2;3;4;6;8;12;24\right\}\)
mà x>10⇒x=\(\left\{12;24\right\}\)
c)\(5.2^2-\left(18:3+2021^0\right)=5.4-6=20-6=14\)
1) \(2^x-15=17\)
\(\Leftrightarrow2^x=32=2^5\)
\(\Rightarrow x=5\)
2) \(\left(7x-11\right)^3=25\cdot5^2+200\)
\(\Leftrightarrow\left(7x-11\right)^3=825\)
\(\Leftrightarrow7x-11=\sqrt[3]{825}\)
\(\Leftrightarrow7x=11+\sqrt[3]{825}\)
\(\Rightarrow x=\frac{11+\sqrt[3]{825}}{7}\)
3) \(\left(x+1\right)^{100}-3\left(x+1\right)^{99}=0\)
\(\Leftrightarrow\left(x+1\right)^{99}\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x+1\right)^{99}=0\\x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=2\end{cases}}\)
4) \(4x+5\left(x+3\right)=105\)
\(\Leftrightarrow9x+15=105\)
\(\Leftrightarrow9x=90\)
\(\Rightarrow x=10\)
5) \(5\cdot\left(x-2\right)+10\left(x+3\right)=170\)
\(\Leftrightarrow5\left[x-2+2\left(x+3\right)\right]=170\)
\(\Leftrightarrow3x+4=34\)
\(\Leftrightarrow3x=30\)
\(\Rightarrow x=10\)
\(3^x+25=26\times2^0+2\times3^0\)
\(3^x+25=26\times1+2\times1\)
\(3^x+25=28\)
\(3^x=28-25\)
\(3^x=3\)
\(x=1\)
3x + 25 = 26 . 20 + 2 . 30
3x + 25 = 26 . 1 + 2 . 1
3x + 25 = 28
3x = 3
x = 1