(13 - 3x)^7 = (13-3x)^4
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15 tháng 3 2017
\(a,\frac{7}{3}x+5=\frac{9}{6}x-7\)
\(\frac{7}{3}x-\frac{9}{6}x=-7-5\)
\(\left(\frac{7}{3}-\frac{9}{6}\right).x=-12\)
\(\frac{5}{6}.x=-12\)
\(x=\left(-12\right):\frac{5}{6}\)
\(x=-\frac{72}{5}\)
\(b,3x+\frac{6}{4}=5-\frac{7x}{6}\)
\(3x+\frac{7x}{6}=5-\frac{6}{4}\)
\(\left(3+\frac{7}{6}\right).x=\frac{7}{2}\)
\(\frac{25}{6}.x=\frac{7}{2}\)
\(x=\frac{7}{2}:\frac{25}{6}\)
\(x=\frac{21}{25}\)
\(c,\frac{12}{5}x+6=\frac{17}{3}x+12\)
\(\frac{12}{5}x-\frac{17}{3}x=12-6\)
\(-\frac{49}{15}x=6\)
\(x=6:\left(-\frac{49}{15}\right)\)
\(x=-\frac{90}{49}\)
\(d,6\left(3x+7\right)=12\left(2x-4\right)\)
\(18x+42=24x-48\)
\(18x-24x=-48-42\)
\(-6x=-90\)
\(x=15\)
\(e,13\left(2x-9\right)=8\left(3x-13\right)\)
\(26x-117=24x-104\)
\(26x-24x=-104+117\)
\(2x=13\)
\(x=\frac{13}{2}\)
BK
1
TA
1 tháng 11 2019
\(\left(\sqrt{7+3x}-4\right)+\left(\sqrt{13-3x}-2\right)+5.\left(\sqrt{\left(7+3x\right)\left(13-3x\right)}-8\right)=0\)
=) \(\frac{7+3x-16}{\sqrt{7+3x}+4}+\frac{13-3x-4}{\sqrt{13-3x}+2}+5.\left(\sqrt{91+18x-9x^2}-8\right)=0\)
=) \(\frac{3\left(x-3\right)}{\sqrt{7+3x}+4}+\frac{3\left(3-x\right)}{\sqrt{13-3x}+2}+\frac{5\left(27+18x-9x^2\right)}{\sqrt{91+18x-9x^2}+8}=0\)
=) \(\frac{3\left(x-3\right)}{\sqrt{7+3x}+4}-\frac{3\left(x-3\right)}{\sqrt{13-3x}+2}-\frac{45\left(x+1\right)\left(x-3\right)}{\sqrt{91+18x-9x^2}+8}=0\)
=) đến đây chắc là tự làm đc rồi
16 tháng 11 2022
Đặt \(\sqrt{7+3x}=a;\sqrt{13-3x}=b\)
=>a+b+5ab=46
=>(a+b)^2=46-5ab
=>a^2+b^2+2ab=2116-460ab+25a^2b^2
=>25a^2b^2-460ab+2116=7+3x+13-3x+2ab
=>25a^2b^2-462ab+2096=0
=>\(\left[{}\begin{matrix}ab=\dfrac{262}{25}\\ab=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(7+3x\right)\cdot\left(13-3x\right)=109.8304\\\left(7+3x\right)\left(13-3x\right)=64\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}91-21x+39x-9x^2=109.8304\\91-21x+39x-9x^2=64\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-9x^2+18x-18.8304=0\\-9x^2+18x+27=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
TT
7
3 tháng 2 2018
3x - 2x = 13 - 19
x = -6
Vậy x = -6
3x + 7 = 4 - 3x
3x + 3x = 7 - 4
6x = 3
x = 3 : 6
x = \(\frac{1}{2}\)
Vậy x = \(\frac{1}{2}\)
5 ( x + 1 ) -3 x = x + 9
5 . x + 5 . 1 - 3x = x + 9
5x + 5 - 3x = x + 9
5x - x - 3x + 5 = 9
4x - 3x = 9 - 5
x = 4
Vậy x = 4
NH
3 tháng 2 2018
A) 3x-2x=13-19
=>. x=-6
Vậy x=-6
b)3x+7=4-3x
=>3x+3x=4-7
=>. 6x=-3
=>. x=-3/6
Vậy x=-3/6
c)5(x+1)-3x=x+9
=>5x+5-3x=x+9
=>. 5x-3x-x=9-5
=>. x=4
Vậy x=4
Chúc bạn học tốt nhé!
HC
0
18 tháng 10 2020
Bài 1:
a) Ta có: \(82-7\left(3x-4\right)=47\)
\(\Leftrightarrow82-21x+28-47=0\)
\(\Leftrightarrow-21x+63=0\)
\(\Leftrightarrow-21x=-63\)
hay x=3(nhận)
Vậy: x=3
b) Ta có: \(97+4\left(5x-7\right)=129\)
\(\Leftrightarrow97+20x-28-129=0\)
\(\Leftrightarrow20x-60=0\)
\(\Leftrightarrow20x=60\)
hay x=3(nhận)
Vậy: x=3
c) Ta có: \(\left(7x-13\right)\cdot27-12=15\)
\(\Leftrightarrow189x-351-12-15=0\)
\(\Leftrightarrow189x-378=0\)
\(\Leftrightarrow189x=378\)
hay x=2(nhận)
Vậy: x=2
d) Ta có: \(\left(2x+3\right)\cdot13+23=140\)
\(\Leftrightarrow26x+39+23-140=0\)
\(\Leftrightarrow26x-78=0\)
\(\Leftrightarrow26x=78\)
hay x=3(nhận)
Vậy: x=3
đ) Ta có: \(52x+8x-5x=70\)
\(\Leftrightarrow55x=70\)
\(\Leftrightarrow x=\frac{70}{55}\)(loại)
Vậy: x∈∅
e) Ta có: \(19x-3x-x=60\)
\(\Leftrightarrow15x=60\)
hay x=4(nhận)
Vậy: x=4
g) Ta có: \(7\left(3x+1\right)-5\left(3x+1\right)=74\)
\(\Leftrightarrow2\left(3x+1\right)=74\)
\(\Leftrightarrow3x+1=37\)
\(\Leftrightarrow3x=36\)
hay x=12(nhận)
Vậy: x=12
h) Ta có: \(5\left(3x-1\right)+7\left(3x-1\right)=96\)
\(\Leftrightarrow12\left(3x-1\right)=96\)
\(\Leftrightarrow3x-1=8\)
\(\Leftrightarrow3x=9\)
hay x=3(nhận)
Vậy: x=3
18 tháng 4 2023
\(1,-\dfrac{4}{7}+\dfrac{2}{3}\times\dfrac{-9}{14}\)
\(=\dfrac{-4}{7}+\dfrac{-18}{42}\)
\(=\dfrac{-4\times6}{7\times6}+\dfrac{-18}{42}\)
\(=\dfrac{-20}{42}+\dfrac{-18}{42}\)
\(=-\dfrac{38}{42}\)
\(=-\dfrac{19}{21}\)
\(2,\dfrac{17}{13}-\left(\dfrac{4}{13}-11\right)\)
\(=\dfrac{17}{13}-\dfrac{4}{13}+11\)
\(=\dfrac{13}{13}+11\)
\(=1+11\)
\(=12\)
\(3,8\dfrac{2}{7}-\left(3\dfrac{4}{9}+4\dfrac{2}{7}\right)\)
\(=\dfrac{58}{7}-\left(\dfrac{31}{9}+\dfrac{30}{7}\right)\)
\(=\dfrac{58}{7}-\dfrac{31}{9}-\dfrac{30}{7}\)
\(=\dfrac{58}{7}-\dfrac{30}{7}-\dfrac{31}{9}\)
\(=\dfrac{28}{7}-\dfrac{31}{9}\)
\(=\dfrac{28\times9}{7\times9}-\dfrac{31\times7}{9\times7}\)
\(=\dfrac{252}{63}-\dfrac{217}{63}\)
\(=\dfrac{35}{63}\)
\(=\dfrac{5}{9}\)
\(5,\left(\dfrac{2}{3}-1\dfrac{1}{2}\right):\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\left(\dfrac{2}{3}-\dfrac{3}{2}\right):\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\left(\dfrac{2\times2}{3\times2}-\dfrac{3\times3}{2\times3}\right):\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\left(\dfrac{4}{6}-\dfrac{9}{6}\right):\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\dfrac{-5}{6}:\dfrac{4}{3}+\dfrac{1}{2}\)
\(=\dfrac{-5}{6}\times\dfrac{3}{4}+\dfrac{1}{2}\)
\(=\dfrac{-15}{24}+\dfrac{1}{2}\)
\(=\dfrac{-15}{24}+\dfrac{1\times12}{2\times12}\)
\(=\dfrac{-15}{24}+\dfrac{12}{24}\)
\(=\dfrac{-3}{24}\)
\(=-\dfrac{1}{8}\)
\(6,\dfrac{-5}{13}+\dfrac{2}{5}+\dfrac{-8}{13}+\dfrac{3}{5}-\dfrac{3}{7}\)
\(=\left(\dfrac{-5}{13}+\dfrac{-8}{13}\right)+\left(\dfrac{2}{5}+\dfrac{3}{5}\right)-\dfrac{3}{7}\)
\(=\dfrac{-13}{13}+\dfrac{5}{5}-\dfrac{3}{7}\)
\(=-1+1-\dfrac{3}{7}\)
\(=-\dfrac{3}{7}\)
\(7,\dfrac{6}{5}\times\dfrac{3}{7}+\dfrac{6}{5}:\dfrac{7}{10}+\dfrac{6}{5}\)
\(=\dfrac{6}{5}\times\dfrac{3}{7}+\dfrac{6}{5}\times\dfrac{10}{7}+\dfrac{6}{5}\)
\(=\dfrac{6}{5}\times\left(\dfrac{3}{7}+\dfrac{10}{7}+1\right)\)
\(=\dfrac{6}{5}\times\left(\dfrac{3}{7}+\dfrac{10}{7}+\dfrac{1\times7}{1\times7}\right)\)
\(=\dfrac{6}{5}\times\left(\dfrac{3}{7}+\dfrac{10}{7}+\dfrac{7}{7}\right)\)
\(=\dfrac{6}{5}\times\dfrac{20}{7}\)
\(=\dfrac{120}{35}\)
\(=\dfrac{24}{7}\)
VD
0
PT
2
11 tháng 5 2021
a)5-(13-3x)=109-(32+109)
5-(13-3x)=-32
13-3x =37
3x =-24
3x =-6
11 tháng 5 2021
a)5-(13-3x)=109-(32+109)
5-13+3x =109-32-109
-8+3x=-32
3x=-32-(-8)
3x=-24
x=-24:3
x=-8
b)|3x-6|+(x-22)=0
(3x-6)+(x-22)=0
3x-6+x-22=0
3x+x=0+6+22
4x=10
x=\(\dfrac{5}{2}\)
c)/7-2x/=-13-5.(-8)
|7-2x|=27
⇒7-2x=27 hoặc 7-2x=-27
x=-10 hoặc x=-17
ta có : (13 - 3x)^7 x 0 = (13-3x)^4 x 0 => (13 - 3x)^7 x 0 - (13-3x)^4 x 0 = 0 = 0 x [(13 - 3x)^7 - (13-3x)^4]
vì : 0 x [(13 - 3x)^7 - (13-3x)^4] và (13 - 3x)^7 x 0 = (13-3x)^4 => (13 - 3^x) x 0 = (13 - 3)^x
kết luận : (13 - 3x)^7 = (13-3x)^4