\(\left(2\times x-15\right)^5=\left(2\times x-15\right)^3\)
Những câu hỏi liên quan
\(\left(2\times x-15\right)^5=\left(2\times x-15\right)^3\)
\(\Leftrightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\Leftrightarrow\left(2x-15\right)^3\cdot\left(\left(2x-15\right)^2-1\right)=0\)
\(2x-15=0\Rightarrow x=\frac{15}{2}\)\(\left(2x-15\right)^2=1\Rightarrow\orbr{\begin{cases}2x-15=1\\2x-15=-1\end{cases}\Rightarrow\orbr{\begin{cases}x=8\\x=7\end{cases}}}\)Vậy, PT có 3 nghiệm x = 7; 15/2; 8.
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tim x
a) 2.\(3^x\) =162
b) \(\left(7\times x-11\right)^3=2^5\times5^2+200\)
c) \(\left(2\times x-15\right)^5=\left(2\times x-15\right)^3\)
a) 2.3x =162
3x =162:2
3x=81
3x=34
vậy x bằng 4
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Tìm xin Z:a. 2timesleft(x-5right)-3timesleft(x-4right)left(-6right)+15timesleft(-3right)b.left(2x-5right)-722c.left(2timesleft|xright|-6right)times1144d.left|x+3right|+left|x+5right|+left|x+9right|4xe.left|x-1right|+left|x-5right|4g.x+left(x+1right)+left(x+2right)+.......+1000
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Tìm \(x\in Z\):
a. \(2\times\left(x-5\right)-3\times\left(x-4\right)=\left(-6\right)+15\times\left(-3\right)\)
b.\(\left(2x-5\right)-7=22\)
c.\(\left(2\times\left|x\right|-6\right)\times11=44\)
d.\(\left|x+3\right|+\left|x+5\right|+\left|x+9\right|=4x\)
e.\(\left|x-1\right|+\left|x-5\right|=4\)
g.\(x+\left(x+1\right)+\left(x+2\right)+.......+100=0\)
a) x=53
b) x=17
c) x=5;x=-5
d) x=17
e) x=5
g) ???
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tim x biet \((x^2-20)\times\left(x^2-15\right)\left(x^2-10\right)\left(x^2-5\right)< 0\)
Tìm \(x\in Z\) biết \(\left(x^3+5\right)\times\left(x^3+10\right)\times\left(x^3+15\right)\times\left(x^3+30\right)
Ta thấy : \(x^3+5\) < \(x^3+10\) < \(x^3+15\) < \(x^3+30\)
Nếu có 1 thừa số âm : \(x^3+5
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Để (x3 + 5) . (x3 + 10) . (x3 + 15) x (x3 + 30) < 0
Mà x3 + 5 < x3 + 10 < x3 + 15 < x3 + 30 nên
<=> x3 + 5 < 0 => x3 < -5 => x \(\le\) -2
hoặc x3 + 5 < 0 và x3 + 10 < 0 và x3 + 15 < 0
=> x3 + 15 < 0 => x3 < -15 => x \(\le-3\)
Vậy \(x\le2\) với \(x\in Z\)
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(x3 + 5)(x3 + 10)(x3 + 15) (x3 + 30 ) < 0
=> trong đó có 3 số âm và 1 số dương hoặc có 3 số dương và 1 số âm
Nhận xét: x3 + 5 < x3 + 10 < x3 + 15 < x3 + 30 . ta có 2 trường hợp sau:
+) TH1: x3 + 5 < x3 + 10 < x3 + 15 < 0 < x3 + 30
=> x3 < -15 và x3 > - 30 => x3 = -29; -28; -27;...; -16 vì x nguyên
Chỉ có x3 = -27 => x = -3 thoả mãn
+) TH2: x3 + 5 < 0< x3 + 10 < x3 + 15 < x3 + 30
=> x3 < -5 và x3 > -10
=> x3 = -9; -8 ; -7; -6 do x nguyên => chỉ có x3 = -8 => x = -2 thoả mãn
Vậy x = -3 hoặc -2
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Xem thêm câu trả lời
1frac{13}{15}times3timesleft(0,5right)^2times3+left(frac{8}{15}-1frac{19}{60}div1frac{23}{24}right)left(-3,2right)timesfrac{-15}{64}+left(0,8-2frac{4}{15}right)div1frac{23}{24} Bài 2 rút gọnfrac{2timesleft(-13right)times9times10}{left(-3right)times4timesleft(-5right)times26}frac{15times8+15times4}{12times3}
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\(1\frac{13}{15}\times3\times\left(0,5\right)^2\times3+\left(\frac{8}{15}-1\frac{19}{60}\div1\frac{23}{24}\right)\)
\(\left(-3,2\right)\times\frac{-15}{64}+\left(0,8-2\frac{4}{15}\right)\div1\frac{23}{24}\)
Bài 2 rút gọn\(\frac{2\times\left(-13\right)\times9\times10}{\left(-3\right)\times4\times\left(-5\right)\times26}\)
\(\frac{15\times8+15\times4}{12\times3}\)
\(1\frac{13}{15}\cdot3\cdot(0,5)^2\cdot3+\left[\frac{8}{15}-1\frac{19}{60}:1\frac{23}{24}\right]\)
\(=\frac{28}{15}\cdot3\cdot0,5\cdot0,5\cdot3+\left[\frac{8}{15}-\frac{79}{60}:\frac{47}{24}\right]\)
\(=\frac{28}{5}\cdot0,25\cdot3+\left[\frac{32}{60}-\frac{79}{60}\cdot\frac{24}{47}\right]\)
\(=\frac{28}{5}\cdot\frac{25}{100}\cdot3+\left[\frac{32}{60}-\frac{158}{235}\right]\)
\(=\frac{28}{5}\cdot\frac{1}{4}\cdot3+\frac{-98}{705}=\frac{7}{5}\cdot1\cdot3+\frac{-98}{705}\)
Đến đây là tính dễ rồi :v
\((-3,2)\cdot\frac{-15}{64}+\left[0,8-2\frac{4}{15}\right]:1\frac{23}{24}\)
\(=\frac{-32}{10}\cdot\frac{-15}{64}+\left[\frac{8}{10}-\frac{34}{15}\right]:\frac{47}{24}\)
\(=\frac{-32\cdot(-15)}{10\cdot64}+\left[\frac{4}{5}-\frac{34}{15}\right]:\frac{47}{24}\)
\(=\frac{-1\cdot(-3)}{2\cdot2}+\frac{4\cdot3-34}{15}:\frac{47}{24}\)
\(=\frac{3}{4}+\frac{-22}{15}:\frac{47}{24}\)
\(=\frac{3}{4}+\frac{-517}{180}=\frac{-191}{90}\)
Bài 2 : \(\frac{2\cdot(-13)\cdot9\cdot10}{(-3)\cdot4\cdot(-5)\cdot26}=\frac{1\cdot(-1)\cdot3\cdot2}{(-1)\cdot2\cdot(-1)\cdot2}=\frac{1\cdot3}{-1\cdot2}=\frac{3}{-2}=\frac{-3}{2}\)
\(\frac{15\cdot8+15\cdot4}{12\cdot3}=\frac{15\cdot(8+4)}{12\cdot3}=\frac{15\cdot12}{12\cdot3}=\frac{15}{3}=5\)
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Tìm X, biết
a,\(5x-16=\left(-36\right)\)
b,\(10-4x=\left(-2\right)\)
c,\(2\times\left(x-5\right)=\left(-10\right)\)
d,\(5\times\left(-16\right)=40+x\)
e,\(4\times\left(-10\right)=15-x\)
f,\(x-15=6+4x\)
a, 5x=-36+16=20
=>x=4
b, 4x=10-(-2)=12
=>x=3
c, x-5=-10 : 2= -5
=> x=0
d, 40+x=-80
=> x=-80-40=-120
e, 15-x=-40
=> x=15-(-40)=55
f, x=6+15+4x=21+4x
=>x-4x=21
=>-3x=21
=> x = -7
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\(2\frac{2}{3}\div\left\{\left[\left(3,72-0,02\times x\right)\frac{10}{37}\right]\div\frac{5}{6}+2,8\right\}-\frac{7}{15}=0,2\)
\(2\frac{2}{3}:\left\{\left[\left(3,72-0,02.x\right)\frac{10}{37}\right]:\frac{5}{6}+2,8\right\}-\frac{7}{15}=0,2\)
\(2\frac{2}{3}:\left\{\left[\left(3,75-0,02.x\right)\frac{10}{37}\right]:\frac{5}{6}+2,8\right\}=\frac{2}{3}\)
\(\left\{\left[\left(3,72-0,02.x\right)\frac{10}{37}\right]:\frac{5}{6}+2,8\right\}=4\)
\(\left[\left(3,72-0,02.x\right)\frac{10}{37}\right]:\frac{5}{6}=\frac{6}{5}\)
\(\left[\left(3,72-0,02.x\right)\frac{10}{37}\right]=1\)
\(\left(3,72-0,02.x\right)=\frac{37}{10}\)
\(0,02.x=0,02\)
\(x=1\)
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\(2\frac{2}{3}:\left\{\left[\left(3,72-0,02.x\right)\frac{10}{37}\right]:\frac{5}{6}+2,8\right\}-\frac{7}{15}=0,2\)
\(\Rightarrow\frac{8}{3}:\left\{\left[\left(\frac{93}{25}-\frac{1}{50}.x\right)\frac{10}{37}\right]:\frac{5}{6}+\frac{14}{5}\right\}-\frac{7}{15}=\frac{1}{5}\)
\(\Rightarrow\left\{\left[\frac{93}{25}-\frac{1}{50}.x\right]:\frac{5}{6}+\frac{14}{5}\right\}-\frac{7}{15}=\frac{8}{3}:\frac{1}{5}=\frac{40}{3}\)
\(\Rightarrow\left[\frac{93}{25}-\frac{1}{50}.x\right]:\frac{5}{6}+\frac{14}{5}=\frac{40}{3}+\frac{7}{15}=\frac{69}{5}\)
\(\Rightarrow\left[\frac{93}{25}-\frac{1}{50}.x\right]:\frac{5}{6}=\frac{69}{5}-\frac{14}{5}=11\)
\(\Rightarrow\frac{93}{25}-\frac{1}{50}.x=11.\frac{5}{6}=\frac{55}{6}\)
\(\Rightarrow\frac{1}{50}.x=\frac{93}{25}-\frac{55}{6}=\frac{-817}{150}\)
\(\Rightarrow x=\frac{-817}{150}:\frac{1}{50}=\frac{-817}{3}\)
Ủng hộ tớ nha m.n?
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\(\left(^{x^2}\times y\right)^{^5}\times\left(x^2\times y^2\right)^7\times\left(x\times y^2\right)^6\times x^3\)
\(\left(x^2.y\right)^5.\left(x^2.y^2\right)^7.\left(x.y^2\right)^6.x^3\)
\(=x^{10}.y^5.x^{14}.y^{14}.x^6.y^{12}.x^3\)
\(=x^{33}.y^{31}\)
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