-2.(x + 6) + 6.(x - 10) = 8 ;
-4.(2x + 9) - (-8x + 3) - (x + 13) = 0 ;
7x.(2 + x) - 7x.(x + 3) = 14.
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b) Ta có: \(5^{x+4}-3\cdot5^{x+3}=2\cdot5^{11}\)
\(\Leftrightarrow2\cdot5^{x+3}=2\cdot5^{11}\)
\(\Leftrightarrow x+3=11\)
hay x=8
c) Ta có: \(2\cdot3^{x+2}+4\cdot3^{x+1}=10\cdot3^6\)
\(\Leftrightarrow18\cdot3^x+12\cdot3^x=10\cdot3^6\)
\(\Leftrightarrow30\cdot3^x=30\cdot3^5\)
Suy ra: x=5
d) Ta có: \(6\cdot8^{x-1}+8^{x+1}=6\cdot8^{19}+8^{21}\)
\(\Leftrightarrow6\cdot\dfrac{8^x}{8}+8^x\cdot8=6\cdot8^{19}+64\cdot8^{19}\)
\(\Leftrightarrow8^x\cdot\dfrac{35}{4}=70\cdot8^{19}\)
\(\Leftrightarrow8^x=8^{20}\)
Suy ra: x=20
(X -10/1994 -1) + (X-8/1996 - 1) + (X-6/1998 - 1)+ (X-4/2000 - 1) + (X-2/2002 - 1) = (X-2002/2 - 1) + (X-2000/4 - 1) + (X-1998/6 - 1) + (X-1996/8 - 1) + (X-1994/10 - 1)
=> x-2004/1994 + x-2004/1996 + x-2004/1998 + x-2004/2000 + x-2004/2002 = x-2004/2 + x-2004/4 + x-2004/6 + x-2004/8 + x-2004/1994
=> x-2004/1994 + x-2004/1996 + x-2004/1998 + x-2004/2000 + x-2004/2002 - x-2004/2 - x-2004/4 - x-2004/6 - x-2004/8 - x-2004/1994 = 0
=> (x - 2004)(1/994 + 1/1996 + 1/1998 + 1/2000 + 1/2002 + 1/2 + 1/4 + 1/6 + 1/8) = 0
Mà (1/994 + 1/1996 + 1/1998 + 1/2000 + 1/2002 + 1/2 + 1/4 + 1/6 + 1/8) \(\ne\)0
=> x - 2004 = 0
=> x = 2004
Vậy x = 2004
6 x 5 = 30
6 x 10 = 60
6 x 2 = 12
6 x 7 = 42
6 x 8 = 48
6 x 3 = 18
6 x 9 = 54
6 x 6 = 36
6 x 4 = 24
`(3xx4xx7)/(5xx3xx4)=7/5`
`(2xx5xx6xx8)/(6xx2xx8xx9)=5/9`
`(4xx5xx6)/(3xx10xx8)= (4xx5xx6)/(3xx5xx2xx4xx2)= 6/(3xx2xx2)= 6/(6xx2)=1/2`
\(a,\dfrac{3\times4\times7}{5\times3\times4}=\dfrac{7}{5}\)
\(b,\dfrac{2\times5\times6\times8}{6\times2\times8\times9}=\dfrac{5}{9}\)
\(c,\dfrac{4\times5\times6}{3\times10\times8}=\dfrac{2}{4}=\dfrac{1}{2}\)
1/ 10(X-7)-8(X+5)=6(-5)+24
10x - 70 - 8x - 40 = -30 +24
2x - 110 = -6
2x = 104
x=52
2/ 8(X-|-7|)-6(X-2)=|-8|.6-50
8(x - 7) - 6(x-2) = 8.6 - 50
8x - 56 - 6x +12 =48 -50
2x - 44 = -2
2x = 42
x=21
3/ 2(4X-8)-7(3+X)=|-4|(3-2)
8x-16 - 21 - 7x = 4.1
x-37=4
x=41
4/ 12(X-4)=6(x-2)-16(X+3)=7|-4|
12x - 48 = 6x - 12 - 16x -48 =7.4
12x - 48 = 28
12x=76
x=19/3
5/ 4(X-5)-7(5-X)+10(5-X)=-3
4x - 20 -35 +7x + 50 -10x = -3
x - 5 = -3
x = -2
Chúc bạn học tốt!
6) 8(x – |-7|) – 6(x – 2) = |-8|.6 – 50
7) -7(5 – x) – 2(x – 10) = 15
8) 4(x – 1) – 3(x – 2) = -|-5|
nhiều quá :((
\(a,2\left(x-5\right)-3\left(x+7\right)=14\)
\(2x-10-3x-21=14\)
\(-x-31=14\)
\(-x=45\)
\(x=45\)
\(b,5\left(x-6\right)-2\left(x+3\right)=12\)
\(5x-30-2x-6=12\)
\(3x-36==12\)
\(3x=48\)
\(x=16\)
\(c,3\left(x-4\right)-\left(8-x\right)=12\)
\(3x-12-8+x=0\)
\(4x-20=0\)
\(4x=20\)
\(x=5\)
Cố nốt nha bn !
cảm ơn, bn nha:)))
mà hình như bạn TOP 3 trả lời câu hỏi pải ko nhỉ???
\(x^2=1\Rightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
\(x^2=3\Rightarrow\left[{}\begin{matrix}x=-\sqrt{3}\\x=\sqrt{3}\end{matrix}\right.\)
\(x^2=5\Rightarrow\left[{}\begin{matrix}x=-\sqrt{5}\\x=\sqrt{5}\end{matrix}\right.\Rightarrow x=-\sqrt{5}\left(vì.x< 0\right)\)
\(x^2=7\Rightarrow\left[{}\begin{matrix}x=-\sqrt{7}\\x=\sqrt{7}\end{matrix}\right.\Rightarrow x=-\sqrt{7}\left(vì.x< 0\right)\)
\(x^2=9\Rightarrow\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\)
\(\left(x-2\right)^2=2\Rightarrow\left[{}\begin{matrix}x-2=-\sqrt{2}\\x-2=\sqrt{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2-\sqrt{2}\\x=2+\sqrt{2}\end{matrix}\right.\)
\(\left(x-4\right)^2=4\Rightarrow\left[{}\begin{matrix}x-2=-2\\x-2=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
\(\left(x-6\right)^2=6\Rightarrow\left[{}\begin{matrix}x-6=-\sqrt{6}\\x-6=\sqrt{6}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6-\sqrt{6}\\x=6+\sqrt{6}\end{matrix}\right.\)
\(\left(x-8\right)^2=8\Rightarrow\left[{}\begin{matrix}x-8=-2\sqrt{2}\\x-8=2\sqrt{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8-2\sqrt{2}\\x=2+2\sqrt{2}\end{matrix}\right.\)
\(\left(x-10\right)^2=10\Rightarrow\left[{}\begin{matrix}x-10=-\sqrt{10}\\x-10=\sqrt{10}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=10-\sqrt{10}\\x=10+\sqrt{10}\end{matrix}\right.\)
\(\left(x-\sqrt{3}\right)^2=3\Rightarrow\left[{}\begin{matrix}x-\sqrt{3}=-\sqrt{3}\\x-\sqrt{3}=\sqrt{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=2\sqrt{3}\end{matrix}\right.\)
\(\left(x-\sqrt{5}\right)^2=5\Rightarrow\left[{}\begin{matrix}x-\sqrt{5}=-\sqrt{5}\\x-\sqrt{5}=\sqrt{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=2\sqrt{5}\end{matrix}\right.\)
+ -2X-12+6X-60=8
<=>4X=80]
<=> X = 20
V.....
+ -8X-36 +8X -3 - X - 13 =0
<=> -X = 52
<=> X = -52
V.....
+ 14X +7\(X^2\)-7\(X^2\)-21X=14
<=> -7X =14
<=> X= -2
V .......