Tim x biết \(2\cdot3^{x+2}+4\cdot3^{x+1}=10\cdot3^6\)
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tim x biet : \(2\cdot3^{x+2}+4\cdot3^{x+1}=10\cdot3^6\)
\(\text{Tìm }x\text{ biết : }\)
\(3^x=3\cdot3^2\cdot3^3\cdot3^4\cdot3^5\cdot...\cdot3^{99}\cdot3^{100}\)
3^x=3^(1+2+3+....+100)
x=1+2+3+..+100
x=(100+1).100/2=5050
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\(3^x=3^1\cdot3^2\cdot...\cdot3^{100}\)
\(3^x=3^{1+2+...+100}\)
\(3^x=3^{5050}\)
Vậy x = 5050
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\(\text{Tìm }x\text{ biết : }\)
\(3^x=3\cdot3^2\cdot3^3\cdot3^4\cdot3^5\cdot...\cdot3^{99}\cdot3^{100}\)
\(3^x=3^1\cdot3^2\cdot3^3\cdot3^4\cdot3^5\cdot...\cdot3^{99}\cdot3^{100}\)
\(3^x=3^{1+2+3+4+5+...+99+100}\)
\(3^x=3^{\left(100+1\right)\cdot\left[\left(100-1\right)\text{ : }1+1\right]\text{ : }2}\)
\(3^x=3^{101\cdot\left(99\text{ }\text{ : }1+1\right)\text{ : }2}\)
\(3^x=3^{101\cdot100\text{ : }2}\)
\(3^x=3^{10100\text{ : }2}\)
\(3^x=3^{5050}\)
\(\Rightarrow\text{ }x=5050\)
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tim x
\(2\cdot3^x=10\cdot3^{12}+8\cdot27^4\)
mk dang can gap giup mk nhe 
\(2.3^x=10.3^{12}+8.27^4\)
\(\Rightarrow\)\(2.3^x=10.3^{12}+2^3.3^{12}\)
\(\Rightarrow\)\(2.3^x=3^{12}\left(10+8\right)\)
\(\Rightarrow\)\(2.3^x=3^{12}.18=3^{12}.2.3.3=3^{14}.2\)
Vậy x = 14
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1.tính c left(1cdot2cdot3cdot4+2cdot3cdot4cdot5+...+18cdot19cdot20cdot21right)-5-10-15-...-200
2. tính d dfrac{1}{1cdot2cdot3}+dfrac{1}{2cdot3cdot4}+...+dfrac{1}{18cdot19cdot20}-dfrac{3}{1cdot2}-dfrac{3}{2cdot3}-...-dfrac{3}{19cdot20}
3.tìm x: 2left(x-3right)-3left(1-2xright)4+4left(1-xright)
4.tìm x: dfrac{x-2}{2}-dfrac{1+x}{3}dfrac{4-3x}{4}-1
giải nhanh giùm tớ nhé. ghi cả lời giả ra kìa
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1.tính c = \(\left(1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot5+...+18\cdot19\cdot20\cdot21\right)-5-10-15-...-200\)
2. tính d = \(\dfrac{1}{1\cdot2\cdot3}+\dfrac{1}{2\cdot3\cdot4}+...+\dfrac{1}{18\cdot19\cdot20}-\dfrac{3}{1\cdot2}-\dfrac{3}{2\cdot3}-...-\dfrac{3}{19\cdot20}\)
3.tìm x: \(2\left(x-3\right)-3\left(1-2x\right)=4+4\left(1-x\right)\)
4.tìm x: \(\dfrac{x-2}{2}-\dfrac{1+x}{3}=\dfrac{4-3x}{4}-1\)
giải nhanh giùm tớ nhé. ghi cả lời giả ra kìa
3/ \(2\left(x-3\right)-3\left(1-2x\right)=4+4\left(1-x\right)\)
\(\Leftrightarrow2x-6-3+6x=4+4-4x\)
\(\Leftrightarrow8x-9=8-4x\)
\(\Leftrightarrow8x+4x=8+9\)
\(\Leftrightarrow12x=17\)
\(\Leftrightarrow x=\dfrac{17}{12}\)
Vậy \(x=\dfrac{17}{12}\)
4/ \(\dfrac{x-2}{2}-\dfrac{1+x}{3}=\dfrac{4-3x}{4}-1\)
\(\Leftrightarrow6\left(x-2\right)-4\left(1+x\right)=3\left(4-3x\right)-12\)
\(\Leftrightarrow6x-12-4-4x=12-9x-12\)
\(\Leftrightarrow6x-4-4x=12-9x\)
\(\Leftrightarrow2x-4=12-9x\)
\(\Leftrightarrow2x+9x=12+4\)
\(\Leftrightarrow11x=16\)
\(\Leftrightarrow x=\dfrac{16}{11}\)
Vậy \(x=\dfrac{16}{11}\)
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Tìm x:
1) \(2\cdot3^x=10\cdot3^{12}+8\cdot27^4\)
2) \(\left(19x+2\cdot5^2\right):14=\left(13-8\right)^2-4^2\)
\(2.3^x=10.3^{12}+8.27^4\)
\(2.3^x=10.531441+8.531441\)
\(2.3^x=\left(10+8\right).531441\)
\(2.3^x=18.531441\)
\(2.3^x=9565938\)
\(3^x=9565938:2\)
\(3^x=4782969\)
\(3^x=3^{14}\)
\(\Rightarrow x=14\)
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1:rút gọn
\(\dfrac{11\cdot3^{22}\cdot3^7-9^{15}}{\left(2\cdot3^{14}\right)^2}\)
2: tìm x
\(\dfrac{3\cdot\left(x-2\right)}{4}-\dfrac{2\cdot\left(1+2x\right)}{3}=1\dfrac{1}{4}-5\cdot\dfrac{\left(1+3x\right)}{6}-\dfrac{x-2}{12}\)
\(\dfrac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\)
\(=\dfrac{11.3^{29}-\left(3^2\right)^{15}}{2^2.3^{28}}\)
\(=\dfrac{11.3^{29}-3^{30}}{2^2.3^{28}}\)
\(=\dfrac{3^{29}\left(11-3\right)}{2^2.3^{28}}\)
\(=\dfrac{3^{29}.2^3}{2^2.3^{28}}\)
\(=\dfrac{3.2}{1.1}=6\)
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Bài 1 : Thực hiện phép tính :
Sdfrac{21^2cdot3^5-4^6cdot9^2}{left(2^2cdot3right)^6+8^4cdot3^5}-dfrac{5^{10}cdot7^3-25^5cdot49^2}{left(125cdot7right)^3cdot5^0cdot14^3}.
Bài 2 : Tìm x :
ɑ) 2^{x-1}+5cdot2^{x-2}dfrac{7}{32}. b) left|xleft(x^2-dfrac{5}{4}right)right|x.
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Bài 1 : Thực hiện phép tính :
\(S=\dfrac{21^2\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3\cdot5^0\cdot14^3}\).
Bài 2 : Tìm x :
ɑ) \(2^{x-1}+5\cdot2^{x-2}=\dfrac{7}{32}\). b) \(\left|x\left(x^2-\dfrac{5}{4}\right)\right|=x\).
a, cho biểu thức P= x-4xy+y .Tinh giá trị của P với \(\left|x\right|=1,5;y=-0,75\)
b, Rút gọn biểu thức: A=\(\dfrac{2^{12}\cdot3^5-4^6\cdot81}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}\)
b, \(A=\dfrac{2^{12}.3^5-4^6.81}{\left(2^2.3\right)^6+8^4.3^5}=\dfrac{2^{12}.3^5-\left(2^2\right)^6.3^4}{2^{12}.3^6+\left(2^3\right)^4.3^5}=\dfrac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}=\dfrac{2^{12}.3^4.\left(3-1\right)}{2^{12}.3^5.\left(3+1\right)}=\dfrac{2}{3.4}=\dfrac{1}{6}\)
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a. Vì | x | = 1,5
\(\Rightarrow\left[{}\begin{matrix}x=1,5\\x=-1,5\end{matrix}\right.\)
Với x = 1,5 ; y = -0,75 thì :
P = 1,5 - 4 . 1,5 . (-0,75) + ( -0,75 )
P = 1,5 - ( -4,5 ) -0,75
P = 6 - 0,75
P = 5,25
Với x = -1,5 ; y = -0,75 thì
P = -1,5 - 4 . ( -1,5 ) .( -0,75 ) + ( -0,75 )
P = -1,5 - 4,5 -0,75
P = -6 - 0,75
P = -6,75
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\(|x+\frac{2017}{2018}|+|2y-6,4|+|z-\frac{2^{19}\cdot3^9\cdot15.2^{18}\cdot9^4}{2^9\cdot3^5\cdot2^{10}\cdot12^{10}}|=0\)
Vì |x + | >= 0 Với mọi x
|2y – 6,4| >= 0 Với mọi y
|z - | >= 0 Với mọi z
ð |x + | + |2y – 6,4| + |z - | >= 0 Với mọi x,y,z
Mà |x + | + |2y – 6,4| + |z - | = 0
Dấu ‘‘=’’ xảy ra khi x + = 0
2y – 6,4 = 0
z - =0
=> x = -
y = 3,2
z = = 33,75
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