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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) \(\left(2x+3\right)^2=\frac{9}{144}\)
\(\Leftrightarrow\left(2x+3\right)^2=\left(\frac{1}{4}\right)^2=\left(-\frac{1}{4}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}2x+3=\frac{1}{4}\\2x+3=\frac{-1}{4}\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=\frac{-11}{4}\\2x=\frac{-13}{4}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{-11}{8}\\x=\frac{-13}{8}\end{cases}}}\)
Vậy ...
b) Ta có: \(\left(3x-1\right)^3=\frac{-8}{27}=\left(\frac{-2}{3}\right)^3\)
\(\Leftrightarrow3x-1=\frac{-2}{3}\Leftrightarrow3x=\frac{1}{3}\Leftrightarrow x=\frac{1}{9}\)
Vậy ....
c) \(x^{10}=25x^8\Leftrightarrow x^{10}:x^8=25\Leftrightarrow x^2=25\Leftrightarrow x=\left\{5;-5\right\}\)
Vậy ...
d) \(\frac{x^7}{81}=27\Leftrightarrow x^7=27.81=2187\)
Mà 37 = 2187 => x7 = 37 => x = 3
Vậy ....
e) \(\frac{x^8}{9}=729\Leftrightarrow x^8=729.9=6561\)
Mà 38 = (-3)8 = 6561
=> x8 = 38 = (-3)8
=> x = {-3;3}
Vậy ...
1. Tìm n, biết:
a) \(\dfrac{-32}{\left(-2\right)^n}=4\)
\(\Rightarrow\dfrac{\left(-2\right)^5}{\left(-2\right)^n}=\left(-2\right)^2\)
\(\Rightarrow\left(-2\right)^n.\left(-2\right)^2=\left(-2\right)^5\)
(-2)n + 2 = (-2)5
n + 2 = 5
n = 5 - 2
n = 3.
b) \(\dfrac{8}{2^n}=2\)
\(\Rightarrow\dfrac{2^3}{2^n}=2\)
\(\Rightarrow\) 2n . 2 = 23
n + 1 = 3
n = 3 - 1
n = 2.
c) \(\left(\dfrac{1}{2}\right)^{2n-1}=\dfrac{1}{8}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^{2n-1}=\left(\dfrac{1}{2}\right)^3\)
2n - 1 = 3
2n = 3 + 1
2n = 4
n = 4 : 2
n = 2.
2. Tính:
a) \(\left(\dfrac{1}{2}\right)^3.\left(\dfrac{1}{4}\right)^2\)
\(=\left(\dfrac{1}{2}\right)^3.\left[\left(\dfrac{1}{2}\right)^2\right]^2\)
\(=\left(\dfrac{1}{2}\right)^3.\left(\dfrac{1}{2}\right)^4\)
\(=\left(\dfrac{1}{2}\right)^7\)
\(=\dfrac{1}{128}\)
b) 273 : 93
= (33)3 : (32)3
= 39 : 36
= 33
= 27
c) 1252 : 253
= (53)2 : (52)3
= 56 : 56
= 1
d) \(\dfrac{27^2.8^5}{6^6.32^3}\)
\(=\dfrac{\left(3^3\right)^2.\left(2^3\right)^5}{6^6.\left(2^5\right)^3}\)
\(=\dfrac{3^6.2^{15}}{6^6.2^{15}}\)
\(=\dfrac{3^6}{6^6}\)
\(=\dfrac{1}{64}.\)
B2 :
b) 27\(^3\): 9\(^3\)= (27:9)\(^3\)= 3\(^3\)
c) 125\(^2\): 25\(^3\)= 15625 : 15625 = 1
Bài1:
Ta có:
a)\(\sqrt{\dfrac{3^2}{5^2}}=\sqrt{\dfrac{9}{25}}=\dfrac{3}{5}\)
b)\(\dfrac{\sqrt{3^2}+\sqrt{42^2}}{\sqrt{5^2}+\sqrt{70^2}}=\dfrac{\sqrt{9}+\sqrt{1764}}{\sqrt{25}+\sqrt{4900}}=\dfrac{3+42}{5+70}=\dfrac{45}{75}=\dfrac{3}{5}\)
c)\(\dfrac{\sqrt{3^2}-\sqrt{8^2}}{\sqrt{5^2}-\sqrt{8^2}}=\dfrac{\sqrt{9}-\sqrt{64}}{\sqrt{25}-\sqrt{64}}=\dfrac{3-8}{5-8}=\dfrac{-5}{-3}=\dfrac{5}{3}\)
Từ đó, suy ra: \(\dfrac{3}{5}=\sqrt{\dfrac{3^2}{5^2}}=\dfrac{\sqrt{3^2}+\sqrt{42^2}}{\sqrt{5^2}+\sqrt{70^2}}\)
Bài 2:
Không có đề bài à bạn?
Bài 3:
a)\(\sqrt{x}-1=4\)
\(\Rightarrow\sqrt{x}=5\)
\(\Rightarrow x=\sqrt{25}\)
\(\Rightarrow x=5\)
b)Vd:\(\sqrt{x^4}=\sqrt{x.x.x.x}=x^2\Rightarrow\sqrt{x^4}=x^2\)
Từ Vd suy ra:\(\sqrt{\left(x-1\right)^4}=16\)
\(\Rightarrow\left(x-1\right)^2=16\)
\(\Rightarrow\left(x-1\right)^2=4^2\)
\(\Rightarrow x-1=4\)
\(\Rightarrow x=5\)
a) \(\frac{1}{81}\times\left(\frac{1}{3}\right)^{-2}\times9\times3^3\)
\(=\frac{3^7}{3^4}\)
\(=3^3\)
b) \(\left(2^5\times4\right)\div\left(2^3\times\frac{1}{16}\right)\)
\(=2^7\div\frac{2^3}{2^{\text{4}}}\)
\(=2^7\div\frac{1}{2}\)
=\(2^6\)
c
\(\sqrt{x}+9=4\)
\(\sqrt{x}=4-9=-5\)
Vì \(\sqrt{x}\) ≥ 0
⇒ Không tồn tại x