(x-3 x 4^2)^3=-64
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a: x^3=7^3
=>x^3=343
=>\(x=\sqrt[3]{343}=7\)
b: x^3=27
=>x^3=3^3
=>x=3
c: x^3=125
=>x^3=5^3
=>x=5
d: (x+1)^3=125
=>x+1=5
=>x=4
e: (x-2)^3=2^3
=>x-2=2
=>x=4
f: (x-2)^3=8
=>x-2=2
=>x=4
h: (x+2)^2=64
=>x+2=8 hoặc x+2=-8
=>x=6 hoặc x=-10
j: =>x-3=2 hoặc x-3=-2
=>x=1 hoặc x=5
k:
9x^2=36
=>x^2=36/9
=>x^2=4
=>x=2 hoặc x=-2
l:
(x-1)^4=16
=>(x-1)^2=4(nhận) hoặc (x-1)^2=-4(loại)
=>x-1=2 hoặc x-1=-2
=>x=3 hoặc x=-1
a) \(\left|x\right|-\frac{7}{6}=\frac{9}{15}\)
=> \(\left|x\right|=\frac{9}{15}+\frac{7}{6}=\frac{53}{30}\)
=> \(\orbr{\begin{cases}x=\frac{53}{30}\\x=-\frac{53}{30}\end{cases}}\)
b) \(\left|x-\frac{4}{3}\right|=\frac{1}{6}\)
=> \(\orbr{\begin{cases}x-\frac{4}{3}=\frac{1}{6}\\x-\frac{4}{3}=-\frac{1}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{7}{6}\end{cases}}\)
c) \(\left|x-\frac{4}{3}\right|-\frac{1}{3}=\frac{1}{2}\)
=> \(\left|x-\frac{4}{3}\right|=\frac{1}{2}+\frac{1}{3}\)
=> \(\left|x-\frac{4}{3}\right|=\frac{5}{6}\)
=> \(\orbr{\begin{cases}x-\frac{4}{3}=\frac{5}{6}\\x-\frac{4}{3}=-\frac{5}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{13}{6}\\x=\frac{1}{2}\end{cases}}\)
d) \(\frac{8}{3}-\left|\frac{7}{9}-x\right|=-\frac{1}{5}\)
=> \(\left|\frac{7}{9}-x\right|=\frac{43}{15}\)
=> \(\orbr{\begin{cases}\frac{7}{9}-x=\frac{43}{15}\\\frac{7}{9}-x=-\frac{43}{15}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{94}{45}\\x=\frac{164}{45}\end{cases}}\)
e) \(\left|x-\left(\frac{1}{4}\right)^2\right|-\frac{25}{64}=0\)
=> \(\left|x-\frac{1}{16}\right|=\frac{25}{64}\)
=> \(\orbr{\begin{cases}x-\frac{1}{16}=\frac{25}{64}\\x-\frac{1}{16}=-\frac{25}{64}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{29}{64}\\x=-\frac{21}{64}\end{cases}}\)
f) \(\left(x-\frac{1}{4}\right)^2+\frac{17}{64}=\frac{21}{32}\)
=> \(\left(x-\frac{1}{4}\right)^2=\frac{25}{64}\)
=> \(\left(x-\frac{1}{4}\right)^2=\left(\frac{5}{8}\right)^2\)
=> \(\orbr{\begin{cases}x-\frac{1}{4}=\frac{5}{8}\\x-\frac{1}{4}=-\frac{5}{8}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{7}{8}\\x=-\frac{3}{8}\end{cases}}\)
\(3\left(x+2\right)^2+\left(2x-3\right)^2-7\left(x-4\right)\left(x+4\right)=64\)
\(\Leftrightarrow3\left(x^2+4x+4\right)+\left(4x^2-12x+9\right)-7\left(x^2-16\right)=64\)
\(\Leftrightarrow3x^2+12x+12+4x^2-12x+9-7x^2+112=64\)
\(\Leftrightarrow12+9+112=64\)(vô lí)
Vậy pt vô nghiệm
TL:
\(\Leftrightarrow3\left(x^2+4x+4\right)+4x^2-6x+9-7x^2+112=64\)
\(\Leftrightarrow6x+133=64\)
\(\Leftrightarrow6x=-69\)
\(\Leftrightarrow x=\frac{-23}{2}\)
Vậy....
a, 410. 230=220.230=250
b,925.274.813= 350.312.312=374
Tương tự các câu khác....
a, 410.230 = (22)10.230 = 220.230 = 250
b, 925.274.813 = (32)25.(33)4.(34)3 = 350.312.312 = 374
c, 2550.1255 = (52)50.(53)5 = 5100.515 = 5115
d, 643.48.164 = (26)3.(22)8.(24)4 = 218.216.216 = 250
e, 38 : 36 = 32
210 : 83 = 210 : (23)3 = 210 : 29 = 2
127 : 67 = (12 : 6)7 = 27
@Dương Tuyết Mai
\(\dfrac{37\cdot5^4}{25^2}=\dfrac{37\cdot5^4}{5^4}=37\\ \dfrac{2^4\cdot2^6\cdot3^8\cdot9^2}{4^4\cdot3^{11}}=\dfrac{2^{10}\cdot3^8\cdot3^4}{2^8\cdot3^{11}}=2^2\cdot3=12\\ \dfrac{3\cdot9^4\cdot9^3}{3^2\cdot9}=\dfrac{3\cdot3^8\cdot3^6}{3^2\cdot3^2}=3^{11}\\ \dfrac{125\cdot5\cdot64-25^3\cdot10\cdot4}{5^7\cdot8}=\dfrac{5^3\cdot5\cdot2^6-5^6\cdot2\cdot5\cdot2^2}{5^7\cdot2^3}=\dfrac{5^4\cdot2^3\left(2^3-5^3\right)}{5^7\cdot2^3}=\dfrac{8-125}{5^3}=\dfrac{-117}{125}\)
Bài 4:
a, \(x^3+12x^2+48x+64=x^3+4x^2+8x^2+32x+16x+64\)
\(=x^2.\left(x+4\right)+8x.\left(x+4\right)+16.\left(x+4\right)\)
\(=\left(x+4\right).\left(x^2+8x+16\right)=\left(x+4\right).\left(x^2+4x+4x+16\right)\)
\(=\left(x+4\right).\left(x+4\right)^2=\left(x+4\right)^3\)(1)
Thay \(x=6\) vào (1) ta được:
\(\left(6+4\right)^3=10^3=1000\)
Vậy...........
b, \(x^3-6x^2+12x-8=x^3-2x^2-4x^2+8x+4x-8\)
\(=x^2.\left(x-2\right)-4x.\left(x-2\right)+4.\left(x-2\right)\)
\(=\left(x-2\right).\left(x^2-4x+4\right)=\left(x-2\right).\left(x^2-2x-2x+4\right)\)
\(=\left(x-2\right).\left(x-2\right)^2=\left(x-2\right)^3\)(2)
Thay \(x=22\) vào (2) ta được:
\(\left(22-2\right)^3=20^3=8000\)
Vậy.............
Chúc bạn học tốt!!!
Bài 2:
a, \(\left(x+9\right)^3=27=3^3\)
\(\Rightarrow x+9=3\Rightarrow x=-6\)
Vậy.........
b, \(8-12x-x^3+6x^2=-64\)
\(\Rightarrow-\left(x^3-6x^2+12x-8\right)=-64\)
\(\Rightarrow x^3-2x^2-4x^2+8x+4x-8=64\)
\(\Rightarrow x^2.\left(x-2\right)-4x.\left(x-2\right)+4.\left(x-2\right)=64\)
\(\Rightarrow\left(x-2\right).\left(x^2-4x+4\right)=64\)
\(\Rightarrow\left(x-2\right).\left(x^2-2x-2x+4\right)=64\)
\(\Rightarrow\left(x-2\right).\left(x-2\right)^2=64\)
\(\Rightarrow\left(x-2\right)^3=4^3\Rightarrow x-2=4\Rightarrow x=6\)
Vậy............
Chúc bạn học tốt!!!
\(5\left(x-1\right)+3=43\)
\(5\left(x-1\right)=40\)
\(x-1=8\)
\(x=9\)
\(\left(1+2+3+...+50\right)+x=5050\)
\(\dfrac{50\left(50+1\right)}{2}+x=5050\)
\(1275+x=5050\)
\(x=3775\)
\(\left(2x+1\right)^3=64\)
\(\left(2x+1\right)^3=4^3\)
\(2x+1=4\)
\(2x=3\)
\(x=1,5\)
`a,`\(2^x -15= 2^4+1\)
`-> 2^x-15=17`
`-> 2^x=17+15`
`-> 2^x=32`
`-> 2^x=2^5`
`-> x=5`
`b,` Có phải đề là \(\dfrac{x+1}{65}+\dfrac{x+2}{64}=\dfrac{x+3}{63}+\dfrac{x+4}{62}\) ?
`=>`\(\dfrac{x+1}{65}+1+\dfrac{x+2}{64}+1=\dfrac{x+3}{63}+1+\dfrac{x+4}{62}+1\)
`=>`\(\dfrac{x+1+65}{65}+\dfrac{x+2+64}{64}-\dfrac{x+3+63}{63}-\dfrac{x+4+62}{62}=0\)
`=>`\(\dfrac{x+66}{65}+\dfrac{x+66}{64}-\dfrac{x+66}{63}-\dfrac{x+66}{62}=0\)
`=>`\(\left(x+66\right)\left(\dfrac{1}{65}+\dfrac{1}{64}-\dfrac{1}{63}-\dfrac{1}{62}\right)=0\)
Mà `1/65+1/64-1/63-1/62 \ne 0`
`-> x+66=0`
`-> x=-66`
a: =>2^x=2^4+16=32
=>x=5
b: Sửa đề: \(\dfrac{x+1}{65}+\dfrac{x+2}{64}=\dfrac{x+3}{63}+\dfrac{x+4}{62}\)
=>\(\left(\dfrac{x+1}{65}+1\right)+\left(\dfrac{x+2}{64}+1\right)=\left(\dfrac{x+3}{63}+1\right)+\left(\dfrac{x+4}{62}+1\right)\)
=>x+66=0
=>x=-66
\(\left(x-3.4^2\right)^3=-64\)
\(\left(x-3.4^2\right)^3=-8^3\)
\(\Rightarrow x-3.4^2=-8\)
\(x-3.16=-8\)
\(x-64=-8\)
\(x=56\)