Tìm x:
\(3.\left(x-9\right)+15=12+243:3^2\)
Những câu hỏi liên quan
tìm x inN1, x^2.x^33^7:3^22, x-18:3123,left(x-18right):2124,x^2165,left(2x+2^4right).5^36,left(3x-48right).63^3.2^2+2^2.3^212^27,1170:left(x-13right)339-3.10^28,left(x-1right).11339,3.left(x-9right)+1512+243:3^2 10,30:left(x-7right)15^9:15^8
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tìm x \(\in\)N
1, \(x^2.x^3=3^7:3^2\)
2, \(x-18:3=12\)
3,\(\left(x-18\right):2=12\)
4,\(x^2=16\)
5,\(\left(2x+2^4\right).5^3\)
6,\(\left(3x-48\right).6=3^3.2^2+2^2.3^2=12^2\)
7,\(1170:\left(x-13\right)=339-3.10^2\)
8,\(\left(x-1\right).11=33\)
9,\(3.\left(x-9\right)+15=12+243:3^2\)
10,\(30:\left(x-7\right)=15^9:15^8\)
1, x^2 . x^3 = 3^7 : 3^2
x^5 = 3^5
=> x = 3
2, x - 18 : 3 = 12
x - 6 = 12
x = 12 + 6
x = 18
3, (x - 18) : 2 = 12
x - 18 = 12 . 2
x - 18 = 24
x = 24 + 18
x = 42
4, x^2 = 16
x^2 = 4^2
=> x = 4
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tìm x biết:
a) \(5^x.\left(5^3\right)^2=625\)
b)\(\left(\dfrac{12}{15}\right)^x=\left(\dfrac{5}{3}\right)^{-5}-\left(-\dfrac{3}{5}\right)^4\)
c)\(\left(-\dfrac{3}{4}\right)^{3x-1}=\dfrac{256}{81}\)
d)\(172x^2-7^9:98^3=2^{-3}\)
tìm x:
\(a,5^x.\left(5^2\right)^3=625\)
\(b,\left(\dfrac{12}{15}\right)^x=\left(\dfrac{5}{4}\right)^{-2}-\left(\dfrac{-3}{5}\right)^4\)
\(c,\left(\dfrac{-3}{4}\right)^{3x-1}=\dfrac{256}{81}\)
\(d,172x^2-7^9:98^3=2^{-3}\)
\(\text{a) 3^{2x} * 9^{2x}=3^{12} b) 7^{3x}= 7^5}:7^3c25x^2=25d\left(x-\frac{\left(2\right)}{3}\right)^5=\frac{1}{243}\)
Tìm x biết : \(\left(\frac{1}{3}\right)^x\left(\frac{1}{9}\right)^x\left(\frac{1}{27}\right)^x\left(\frac{1}{81}\right)^x\left(\frac{1}{243}\right)^x=\left(-\frac{1}{3}\right)^{30}\)
bài 1: Tìm x,y biết rằng: x+(-frac{31}{12})^2left(frac{49}{12}right)^2-xy^2 bài 2: tìm x biết:a.5^x.left(5^3right)^2625 b.left(frac{12}{25}right)^xleft(frac{5}{3}right)^{-2}-left(-frac{3}{5}right)^4 c.left(-frac{3}{4}right)^{3x-1}frac{256}{81} d.172x^2-7^9:98^32^{-3}Bài 3: Tìm x varepsilonN biết:a.8 2^xle2^9times2^{-5} b.27 81^3:3^x 243 left(frac{2}{5}right)^xleft(frac{5}{2}right)^{-3}timesleft(-frac{2}{5}right)^2c.
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bài 1: Tìm x,y biết rằng:
\(x+(-\frac{31}{12})^2=\left(\frac{49}{12}\right)^2-x=y^2\)
bài 2: tìm x biết:
a.\(5^x.\left(5^3\right)^2=625\) b.\(\left(\frac{12}{25}\right)^x=\left(\frac{5}{3}\right)^{-2}-\left(-\frac{3}{5}\right)^4\) c.\(\left(-\frac{3}{4}\right)^{3x-1}=\frac{256}{81}\)
d.\(172x^2-7^9:98^3=2^{-3}\)
Bài 3: Tìm x \(\varepsilon\)N biết:
a.\(8< 2^x\le2^9\times2^{-5}\) b.\(27< 81^3:3^x< 243\) \(\left(\frac{2}{5}\right)^x>\left(\frac{5}{2}\right)^{-3}\times\left(-\frac{2}{5}\right)^2\)c.
Bài 1:
Ta có: \(x+\left(-\frac{31}{12}\right)^2=\left(\frac{49}{12}\right)^2-x\)
\(\Leftrightarrow2x=\frac{1440}{144}=10\)
\(\Rightarrow x=5\)
Khi đó: \(y^2=\left(\frac{49}{12}\right)^2-5=\frac{1681}{144}\)
=> \(\hept{\begin{cases}y=\frac{41}{12}\\y=-\frac{41}{12}\end{cases}}\)
3,2.frac{15}{16}-left(75%+frac{2}{7}right):left(-1frac{1}{28}right)left(0,25+12,5-frac{5}{16}right):left[12-frac{7}{12}:left(frac{3}{8}-frac{1}{12}right)right]left(frac{-3}{5}+frac{5}{11}right):frac{-3}{7}+left(frac{-2}{5}+frac{6}{5}right):frac{-3}{7}14,5-frac{8}{9}:left(35-34frac{8}{9}right).frac{9}{8}1frac{1}{15}-left(frac{1}{15}+frac{4}{9}:frac{-2}{3}-frac{28}{16}.frac{6}{35}right)-frac{3}{10}Tìm x left(4,5-2xright)left(-3frac{2}{3}right)frac{11}{15}backslash34-xbackslashleft(-3right)^4left(4...
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\(3,2.\frac{15}{16}-\left(75\%+\frac{2}{7}\right):\left(-1\frac{1}{28}\right)\)
\(\left(0,25+12,5-\frac{5}{16}\right):\left[12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right]\)
\(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)
\(14,5-\frac{8}{9}:\left(35-34\frac{8}{9}\right).\frac{9}{8}\)
\(1\frac{1}{15}-\left(\frac{1}{15}+\frac{4}{9}:\frac{-2}{3}-\frac{28}{16}.\frac{6}{35}\right)-\frac{3}{10}\)
Tìm x
\(\left(4,5-2x\right)\left(-3\frac{2}{3}\right)=\frac{11}{15}\)
\(\backslash34-x\backslash=\left(-3\right)^4\)
\(\left(4x^2-1\right)\left(\text{\x}\backslash-\frac{2}{3}\right)=0\)
\(\frac{3}{5}x-\frac{1}{2}\)\(x=\frac{-7}{20}\)
\(=\frac{16}{5}.\frac{15}{16}-\left(\frac{3}{4}+\frac{2}{7}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{21}{28}+\frac{8}{28}\right):\left(\frac{-29}{28}\right)\)
\(=3-\left(\frac{29}{28}\right).\left(\frac{-28}{29}\right)\)
\(=3-\left(-1\right)\)
\(=4\)
b) \(=\left(\frac{1}{4}+\frac{25}{2}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3}{8}-\frac{1}{12}\right)\right)\)
\(=\left(\frac{4}{16}+\frac{200}{16}-\frac{5}{16}\right):\left(12-\frac{7}{12}:\left(\frac{3.3}{2.3.4}-\frac{2}{2.3.4}\right)\right)\)
\(=\left(\frac{199}{16}\right):\left(12-\frac{7}{12}:\left(\frac{9}{24}-\frac{2}{24}\right)\right)\)
\(=\frac{199}{16}:\left(12-\frac{7}{12}.\frac{24}{7}\right)\)
\(=\frac{199}{16}:\left(12-2\right)\)
\(=\frac{199}{16}:10\)
\(=\frac{199}{160}\)
c) \(\left(\frac{-3}{5}+\frac{5}{11}\right):\frac{-3}{7}+\left(\frac{-2}{5}+\frac{6}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-33}{55}+\frac{25}{55}\right):\frac{-3}{7}+\left(\frac{4}{5}\right):\frac{-3}{7}\)
\(\left(\frac{-8}{55}\right).\frac{-7}{3}+\frac{4}{5}.\frac{-7}{3}\)
\(\frac{-7}{3}\left(\frac{-8}{55}+\frac{4}{5}\right)\)
\(\frac{-7}{3}.\frac{36}{55}=\frac{-84}{55}\)
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nhớ làm giúp mình nhá mai mình phải đi học r :<
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Xem thêm câu trả lời
Tìm x
\(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=15\)
\(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)^2=15\)
\(\Leftrightarrow x^3-9x^2+27x-27-\left(x^3-3^3\right)+9\left(x+1\right)^2=15\)
\(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+9.\left(x^2+2x+1\right)=15\)
\(\Leftrightarrow-9x^2+27x+9x^2+18x+9=15\)
\(\Leftrightarrow45x=6\Leftrightarrow x=\frac{2}{15}\)
Vậy: \(S=\left\{\frac{2}{15}\right\}\)
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pt <=> \(\left(x-3\right)^3-\left(x^3-27\right)+9\left(x+1\right)^2=15\)
<=> \(x^3-3x^2.3+3x.3^2-27-x^3+27+9x^2+18x+9=15\)
<=> \(45x=6\)
<=> \(x=\frac{6}{45}=\frac{2}{15}\)
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Rút gọn: a)\(\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}.12^{10}}\)
b)\(\frac{\left(-\frac{1}{2}\right)^3-\left(\frac{3}{4}\right)^3.\left(-2\right)^2}{2.\left(-1\right)^5+\left(\frac{3}{4}\right)^2-\frac{3}{8}}\)
Tìm x
a)\(3^{x+1}=9^x\)
b)\(2^{3x+2}=4^{x+5}\)
c)\(3^{2x-1}=243\)
Tìm x:8) 1-left(x-6right)4left(2-2xright)9)left(3x-2right)left(x+5right)010)left(x+3right)left(x^2+2right)011)left(5x-1right)left(x^2-9right)012)xleft(x-3right)+3left(x-3right)013)xleft(x-5right)-4x+20014)x^2+4x-50
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Tìm \(x\):
\(8\)) \(1-\left(x-6\right)=4\left(2-2x\right)\)
\(9\))\(\left(3x-2\right)\left(x+5\right)=0\)
\(10\))\(\left(x+3\right)\left(x^2+2\right)=0\)
\(11\))\(\left(5x-1\right)\left(x^2-9\right)=0\)
\(12\))\(x\left(x-3\right)+3\left(x-3\right)=0\)
\(13\))\(x\left(x-5\right)-4x+20=0\)
\(14\))\(x^2+4x-5=0\)
\(8,1-\left(x-6\right)=4\left(2-2x\right)\)
\(\Leftrightarrow1-x+6=8-8x\)
\(\Leftrightarrow-x+8x=8-1-6\)
\(\Leftrightarrow7x=1\)
\(\Leftrightarrow x=\dfrac{1}{7}\)
\(9,\left(3x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)
\(10,\left(x+3\right)\left(x^2+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\varnothing\end{matrix}\right.\)
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`8)1-(x-5)=4(2-2x)`
`<=>1-x+5=8-6x`
`<=>5x=2<=>x=2/5`
`9)(3x-2)(x+5)=0`
`<=>[(x=2/3),(x=-5):}`
`10)(x+3)(x^2+2)=0`
Mà `x^2+2 > 0 AA x`
`=>x+3=0`
`<=>x=-3`
`11)(5x-1)(x^2-9)=0`
`<=>(5x-1)(x-3)(x+3)=0`
`<=>[(x=1/5),(x=3),(x=-3):}`
`12)x(x-3)+3(x-3)=0`
`<=>(x-3)(x+3)=0`
`<=>[(x=3),(x=-3):}`
`13)x(x-5)-4x+20=0`
`<=>x(x-5)-4(x-5)=0`
`<=>(x-5)(x-4)=0`
`<=>[(x=5),(x=4):}`
`14)x^2+4x-5=0`
`<=>x^2+5x-x-5=0`
`<=>(x+5)(x-1)=0`
`<=>[(x=-5),(x=1):}`
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\(11,=>\left[{}\begin{matrix}5x-1=0\\x^2-9=0\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=3\\x=-3\end{matrix}\right.\\ 12,=>\left(x+3\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x+3=0\\x-3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\\ 13,=>x\left(x-5\right)-4\left(x-5\right)=0\\ =>\left(x-4\right)\left(x-5\right)=0\\ =>\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
\(14,=>x^2+5x-x-5=0\\ =>x\left(x+5\right)-\left(x+5\right)=0\\ =>\left(x-1\right)\left(x+5\right)=0\\ =>\left[{}\begin{matrix}x-1=0\\x+5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)
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