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\(\frac{9^5.16^4}{27^3.^{ }8^4}=\frac{\left(3^2\right)^5.\left(2^4\right)^4}{\left(3^3\right)^3.\left(2^3\right)^4}=\frac{3^{10}.2^{16}}{3^9.2^{12}}=3.2^4=3.16=48\)
27x . 9x = 927 : 81
=> (33)x . (32)x = (32)27 : 34
=> 33x . 32x = 354 : 34
=> 35x = 350
=> 5x = 50
=> x = 50 : 5
=> x = 10
1.\(45^{10}.5^{30}=45^{10}.125^{10}=\left(45.125\right)^{10}=5625^{10}\)
2.a. \(\left(2x-1\right)^3=-8\Leftrightarrow\left(2x-1\right)^3=\left(-2\right)^3\)
\(\Leftrightarrow2x-1=-2\Leftrightarrow x=-\frac{1}{2}\)
b.\(\left(x+\frac{1}{2}\right)^2=\frac{1}{16}\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=\frac{1}{4}\\x+\frac{1}{2}=-\frac{1}{4}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{4}\\x=-\frac{3}{4}\end{cases}}\)
c. \(\left(2x+3\right)^2=\frac{9}{121}\Leftrightarrow\orbr{\begin{cases}2x+3=\frac{3}{11}\\2x+3=-\frac{3}{11}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{15}{11}\\x=-\frac{18}{11}\end{cases}}\)
d.\(\left(3x-1\right)^3=-\frac{8}{27}=\left(-\frac{2}{3}\right)^3\)
\(\Leftrightarrow3x-1=-\frac{2}{3}\Leftrightarrow x=\frac{1}{9}\)
4.
a.\(99^{20}=\left(99^2\right)^{10}=9801^{10}\)
Do \(9801^{10}< 9999^{10}\Rightarrow99^{20}< 9999^{10}\)
b.\(3^{4000}=\left(3^2\right)^{2000}=9^{2000}\)
\(\Rightarrow3^{4000}=9^{2000}\)
c.\(2^{332}=\left(2^3\right)^{110}.2^2=8^{110}.4\)
\(3^{223}=\left(3^2\right)^{110}.3^3=\left(3^2\right)^{110}.9=9^{110}.9\)
Ta thấy \(4.8^{110}< 9.9^{110}\)
Vậy \(2^{332}< 3^{223}\)
Đúng vì:
log_3(x+1) = 2 - log_2(x)
= 2 log_2(2) - log_2(x)
= log_2(2^2) - log_2(x)
= log_2(4/x)
log (x+1)/log 3 = log (4/x)/log 2
log (x+1)/log (4/x) = log 3/log 2
Rearrange:
log_2 3 = log_4x (x+1)
so 4x = 2 and (x+1) = 3
x = 4/2 = 2 or x = 3-1 = 2
P/s :Ko chắc đâu nhé
x2 : y2=16/9 =>x2/16=y2/9.
áp dụng tính chất của dãy tỉ số bằng nhau,ta có:x2/16=y2/9=x2+y2/16+9=100/25=4
=>x2=4.16=64 .Mà x là số nguyên dương nên x=8
=>y2=4.9=36 .Mà y là số nguyên dương nên y=6
( x : y ) 2 =16/9
=> x2 : y 2 = 16/9
=> x 2 = 16/9. y 2
=> 16/9.y 2 + y 2 =100
=> y 2.(16/9 +1 ) = 100
=> 25/9 . y 2 = 100
=> y 2 = 36
=> y = 6, y= -6
=> x2 + 36 = 100
=> x2 =64
=> x = 8 , x = -8
b) Tính
\(A=\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}\)
\(=\frac{\left(2^4\right)^3.3^{10}+2^3.3.5.2^9.3^9}{\left(2^2\right)^6.3^{12}+2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}.\left(2.3+1\right)}\)
\(=\frac{2.6}{3.7}=\frac{12}{21}=\frac{4}{7}\)
Vậy : \(A=\frac{4}{7}\)
94 . 86 / 610. 163
= (32)4 . (23)8 / ( 3.2)10 . (24)3
=38 . 224 / 310 . 210 . 212
= 22 / 32
= 4 / 9
= 0.(4)
Chúc bạn học tốt nhé!!!!
94 . 86 / 610 . 163
= 38 . 218 / 320 . 212
= 1. 26 / 312 . 1
Tự làm tiếp nha bạn!
\(1)-4x\left(x-5\right)-2x\left(8-2x\right)=-3\)
\(\Rightarrow-4x^2-\left(-20x\right)-16x+4x^2=-3\)
\(\Rightarrow20x-14x=-3\)
\(\Rightarrow6x=-3\)
\(\Rightarrow x=-\dfrac{1}{2}\)
Vậy \(x=-\dfrac{1}{2}\)
\(2)\) Theo bài ra, ta có: \(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\) và \(x^2+y^2+z^2=14\)
\(\Rightarrow\dfrac{x^3}{2^3}=\dfrac{y^3}{4^3}=\dfrac{z^3}{6^3}\)
\(\Rightarrow\left(\dfrac{x}{2}\right)^3=\left(\dfrac{y}{4}\right)^3=\left(\dfrac{z}{6}\right)^3\)
\(\Rightarrow\sqrt[3]{\left(\dfrac{x}{2}\right)^3}=\sqrt[3]{\left(\dfrac{y}{4}\right)^3}=\sqrt[3]{\left(\dfrac{z}{6}\right)^3}\)
\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{6}\)
\(\Rightarrow\left(\dfrac{x}{2}\right)^2=\left(\dfrac{y}{4}\right)^2=\left(\dfrac{z}{6}\right)^2\)
\(\Rightarrow\dfrac{x^2}{2^2}=\dfrac{y^2}{4^2}=\dfrac{z^2}{6^2}\)
\(\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{16}=\dfrac{z^2}{36}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{x^2}{4}=\dfrac{y^2}{16}=\dfrac{z^2}{36}=\dfrac{x^2+y^2+z^2}{4+16+36}=\dfrac{14}{56}=\dfrac{1}{4}\)
Suy ra:
\(+)\dfrac{x^2}{4}=\dfrac{1}{4}\Rightarrow x^2=\dfrac{1}{4}.4=1=\left(\pm1\right)^2\Rightarrow x=\pm1\)
\(+)\dfrac{y^2}{16}=\dfrac{1}{4}\Rightarrow y^2=\dfrac{1}{16}.4=\dfrac{1}{4}=\left(\pm\dfrac{1}{2}\right)^2\Rightarrow y=\pm\dfrac{1}{2}\)
\(+)\dfrac{z^2}{36}=\dfrac{1}{4}\Rightarrow z^2=\dfrac{1}{36}.4=\dfrac{1}{9}=\left(\pm\dfrac{1}{3}\right)^2\Rightarrow z=\pm\dfrac{1}{3}\)
Vậy \(\left(x;y;z\right)\in\left\{\left(-1;-\dfrac{1}{2};-\dfrac{1}{3}\right);\left(1;\dfrac{1}{2};\dfrac{1}{3}\right)\right\}\)
1. S = { 3;4 }
2. S={ -2; 1}
3. S={\(\frac{1}{2}\) ; 2;-2}
4.S={\(\frac{4}{3}\) ;2}
S la tap ngo nhek , xin k nao
\(\Leftrightarrow\left(\dfrac{4}{3}\right)^{150}:x=\left(-\dfrac{4}{3}\right)^{135}\)
\(\Leftrightarrow x=\left(\dfrac{4}{3}\right)^{150}:\left(-\dfrac{4}{3}\right)^{135}=-\left(\dfrac{4}{3}\right)^{15}\)