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a: \(\Leftrightarrow4^x\left(\dfrac{3}{2}+\dfrac{5}{3}\cdot4^2\right)=4^8\left(\dfrac{3}{2}+\dfrac{5}{3}\cdot4^2\right)\)
=>4^x=4^8
=>x=8
b: \(\Leftrightarrow2^x\cdot\dfrac{1}{2}+2^x\cdot2=2^{10}\left(2^2+1\right)\)
=>2^x=2^11
=>x=11
c: =>1/6*6^x+6^x*36=6^15(1+6^3)
=>6^x=6*6^15
=>x=16
d: \(\Leftrightarrow8^x\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)=8^9\left(\dfrac{5}{3}\cdot8^2-\dfrac{3}{5}\right)\)
=>x=9
1.
\(-3x^5y^4+3x^2y^3-7x^2y^3+5x^5y^4\)
\(=(-3x^5y^4+5x^5y^4)+(3x^2y^3-7x^2y^3)\)
\(=2x^5y^4-4x^2y^3\)
2.
\(\frac{1}{2}x^4y-\frac{3}{2}x^3y^4+\frac{5}{3}x^4y-x^3y^4\)
\(=(\frac{1}{2}x^4y+\frac{5}{3}x^4y)-(\frac{3}{2}x^3y^4+x^3y^4)\)
\(=\frac{13}{6}x^4y-\frac{5}{2}x^3y^4\)
3.
\(5x-7xy^2+3x-\frac{1}{2}xy^2\)
\(=(5x+3x)-(7xy^2+\frac{1}{2}xy^2)\)
\(=8x-\frac{15}{2}xy^2\)
4.
\(\frac{-1}{5}x^4y^3+\frac{3}{4}x^2y-\frac{1}{2}x^2y+x^4y^3\)
\(=(\frac{-1}{5}x^4y^3+x^4y^3)+(\frac{3}{4}x^2y-\frac{1}{2}x^2y)\)
\(=\frac{4}{5}x^4y^3+\frac{1}{4}x^2y\)
5.
\(\frac{7}{4}x^5y^7-\frac{3}{2}x^2y^6+\frac{1}{5}x^5y^7+\frac{2}{3}x^2y^6\)
\(=(\frac{7}{4}x^5y^7+\frac{1}{5}x^5y^7)+(-\frac{3}{2}x^2y^6+\frac{2}{3}x^2y^6)\)
\(=\frac{39}{20}x^5y^7-\frac{5}{6}x^2y^6\)
6.
\(\frac{1}{3}x^2y^5(-\frac{3}{5}x^3y)+x^5y^6=(\frac{1}{3}.\frac{-3}{5})(x^2.x^3)(y^5.y)+x^5y^6\)
\(=\frac{-1}{5}x^5y^6+x^5y^6=\frac{4}{5}x^5y^6\)
a, ( x-1)3= -27
=> x - 1 = -3
=> x = -2
b, ( 2x - 1)2=25
=> 2x - 1 = 5 hoặc 2x - 1 = -5
=> 2x = 6 hoặc 2x = -4
=> x = 3 hoặc x = -2
c, ( x - 3/4)2= ( 1/2)6
=> (x - 3/4)^2 = 1/64
=> x - 3/4 = 1/8 hoặc x - 3/4 = -1/8
=> x = 7/8 hoặc x = 5/8
d, 2 x + 2 x +2 = 80
=> 2^x + 2^x.4 = 80
=> 2^x(1 + 4) = 80
=> 2^x.5 = 80
=> 2^x = 16
=> x = 4
e, 4x + 4 x + 3 = 4160
=> 4^x(1 + 64) = 4160
=> 4^x.65 = 4160
=> 4^x = 64
=> x = 3
1) Tính
a) 253 : 52 = (52)3 : 52 = 56 : 52 = 54 = 625
\(b)\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{9}{49}\right)^6=\left(\dfrac{3}{7}\right)^{21}:\left[\left(\dfrac{3}{7}\right)^2\right]^6=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{3}{7}\right)^{12}=\left(\dfrac{3}{7}\right)^9\) d) 9 . 32 . \(\dfrac{1}{81}\) . 32 = 32 . 32 . \(\dfrac{1}{3^4}\) . 32 = 9
2) Tìm x thuộc Q, biết:
a) 3x + 2 = 27
=> 3x + 2 = 33
x + 2 = 3
x = 3 - 2
x = 1
b) \(\left(\dfrac{1}{2}x-3\right)^4=81\)
\(\Rightarrow\left(\dfrac{1}{2}x-3\right)^4=3^4\)
\(\dfrac{1}{2}x-3=3^{ }\)
\(\dfrac{1}{2}x=3+3\)
\(\dfrac{1}{2}x=9\)
\(x=9:\dfrac{1}{2}\)
\(x=18\)
c) \(\left(x-\dfrac{1}{2}\right)^3=-27\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^3=\left(-3\right)^3\)
\(x-\dfrac{1}{2}=-3\)
\(x=-3+\dfrac{1}{2}\)
\(x=\dfrac{-5}{2}\)
d) 5 . 5x + 1 = 125
5x + 1 = 125 : 5
5x + 1 = 25
5x + 1 = 52
x + 1 = 2
x = 2 - 1
x = 1.
1: \(=\dfrac{3}{4}+\dfrac{5}{4}\cdot\dfrac{8}{3}-\dfrac{1}{4}\cdot\dfrac{5}{6}=\dfrac{3}{4}+\dfrac{10}{3}-\dfrac{5}{24}\)
\(=\dfrac{18}{24}+\dfrac{80}{24}-\dfrac{5}{24}=\dfrac{93}{24}=\dfrac{31}{8}\)
2: \(=\left(7+\dfrac{23}{27}-\dfrac{23}{27}\right)+\left(\dfrac{11}{25}+\dfrac{14}{25}\right)+3.25\)
\(=7+1+3.25=8+3.25=11.25\)
3: \(=\left(\dfrac{1}{9}\cdot9\right)^{2005}-4^2=1-16=-15\)
4: \(=2\cdot\dfrac{9}{4}-\dfrac{7}{2}=\dfrac{9}{2}-\dfrac{7}{2}=1\)
5: \(=\dfrac{15}{2}\cdot\dfrac{-3}{5}+\dfrac{5}{2}\cdot\dfrac{-3}{5}=\dfrac{-3}{5}\cdot\left(\dfrac{15}{2}+\dfrac{5}{2}\right)=\dfrac{-3}{5}\cdot10=-6\)
6: \(=\left(\dfrac{6}{10}+\dfrac{5}{10}\right)^2=\left(\dfrac{11}{10}\right)^2=\dfrac{121}{100}\)
7: \(=\dfrac{1}{2}\cdot\dfrac{-7}{2}=\dfrac{-7}{4}\)
Ta có : 5(x - 2)(x + 3) = 1
=> (5x - 10)(x + 3) = 1
=> 5x2 - 10x + 15x - 30 = 1
=> 5x2 - 5x - 30 = 1
=> 5x(x - 1) = 31
=> x(x - 1) = 31/5 (chịu)
Bài 2
\(a,\left(x-3\right)^2=9\Leftrightarrow\left(x-3\right)^2=3^2\Leftrightarrow x-3=3\Leftrightarrow x=6\)
\(b,\left(\frac{1}{2}+x\right)^2=16\Leftrightarrow\left(\frac{1}{2}+x\right)^2=4^2\Leftrightarrow\frac{1}{2}+x=4\Leftrightarrow x=\frac{7}{2}\)
a)Viết dưới dạng phân số rồi sử dụng tích chéo ý
b)\(\frac{-1}{7}.2^3-2x:1\frac{4}{3}=-2^{x-1}\)
\(\Rightarrow\frac{-8}{7}-2x:\frac{7}{3}=-2^{x-1}\)
\(\Rightarrow\frac{-8}{7}-\frac{6x}{7}=-2^{x-1}\)
\(\Rightarrow\frac{-8-6x}{7}=\frac{2^{x-1}}{-1}\)
\(\Rightarrow-1\left(-8-6x\right)=7.2^{x-1}\)
\(\Rightarrow6x+8=7.2^{x-1}\)
.........
a: \(\dfrac{\left(-1\right)^2}{2^2}=\dfrac{1}{4};\left(-\dfrac{1}{2}\right)^2=\dfrac{1}{4}\)
Do đó: \(\dfrac{\left(-1\right)^2}{2^2}=\left(-\dfrac{1}{2}\right)^2\)
b: \(\dfrac{3^3}{5^3}=\left(\dfrac{3}{5}\right)^3< \dfrac{3}{5}\)(do \(0< \dfrac{3}{5}< 1\))
d: \(\left(\dfrac{3}{4}\right)^7:\left(\dfrac{3}{4}\right)^3=\left(\dfrac{3}{4}\right)^4\)
Vì \(0< \dfrac{3}{4}< 1\)
nên \(\left(\dfrac{3}{4}\right)^4< \left(\dfrac{3}{4}\right)^2\)
=>\(\left(\dfrac{3}{4}\right)^7:\left(\dfrac{3}{4}\right)^3< \left(\dfrac{3}{4}\right)^2\)
e: \(\left(0,5\right)^6:\left(0,5\right)^2=\left(0,5\right)^{6-2}=\left(0,5\right)^4=\left(0,5\right)^{2\cdot2}=\left[\left(0,5\right)^2\right]^2\)
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