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a) ( x - 8 ) ( x3 +8 ) = 0
=> x - 8 = 0 hoặc x3 + 8 = 0
+) x -8 = 0
x = 8
+) x3 + 8 = 0
x3 =-8 ( vô lí )
Vậy x = 8
a) \(3^x=\frac{3^8}{3^9}=\frac{1}{3}=3^{-1}\)
\(\Rightarrow x=-1\)
Vậy x = -1
b) \(\frac{3^5}{3^x}=3^{10}\)
\(\Rightarrow3^5:3^x=3^{10}\)
\(\Rightarrow3^x=3^5:3^{10}\)
\(\Rightarrow3^x=\frac{1}{3^5}\)
\(\Rightarrow3^x=3^{-5}\)
\(\Rightarrow x=-5\)
Vạy x = -5
c) \(\left(-5\right)^x=\frac{25^{10}}{\left(-5\right)^{17}}\)
\(\Rightarrow\left(-5\right)^x=\frac{5^{20}}{\left(-5\right)^{17}}\)
\(\Rightarrow\left(-5\right)^x=\left(-5\right)^3\)
\(\Rightarrow x=3\)
Vậy x = 3
1) Tìm số nguyên x, biết :
a) 3x = 94/ 273
3x = 1/3
3x = 3-1
=> x = -1
b) 3x = 98 / 273 . 812
3x = 37.38
3x = 315
=> x = 15
c) 2x - 3 / 410 = 83
2x - 3 = 83.410
2x - 3 = 226
=> x - 3 = 26
=> x = 29
d) 22x - 3 / 410 = 83 . 165
22x - 3 / 410 = 269
22x - 3 = 269 . 410
22x - 3 = 289
=> 2x - 3 = 89
2x = 91
x = 91/2
e) 35 / 3x = 310
3x = 35 : 310
3x = 3-5
=> x = -5
TH1: a+b+c khác 0
\(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}\)
\(\Rightarrow2+\frac{a+b-c}{c}=2+\frac{b+c-a}{a}=2+\frac{c+a-b}{b}\)
\(\Rightarrow\frac{a+b+c}{c}=\frac{a+b+c}{a}=\frac{a+b+c}{b}\)
\(\Rightarrow a=b=c\)
thay a=b=c vào B ta có:
\(B=\left(1+\frac{a}{a}\right)\cdot\left(1+\frac{a}{a}\right)\cdot\left(1+\frac{a}{a}\right)=2\cdot2\cdot2=8\)
TH2: a+b+c=0
=> c=-a-b
=>a=-b-c
=>b=-a-c
thay a,b,c vào B ta có:
\(B=\left(1+\frac{-\left(a+c\right)}{a}\right)\cdot\left(1+\frac{-\left(b+c\right)}{c}\right)\cdot\left(1+\frac{-\left(a+b\right)}{b}\right)\)
\(B=\left(-\frac{c}{a}\right)\cdot\left(-\frac{b}{c}\right)\cdot\left(-\frac{a}{b}\right)=-1\)
p/s: th2 ko chắc nhá
a, x : (-1/2)^3 = -1/2
=> x : (-1/8) = -1/2
=> x = 4
vậy_
b, (3/4)^5.x = (3/4)^7
=> x = (3/4)^7 : (3/4)^5
=> x = (3/4)^2
=> x = 9/16
vậy-
c, (3/5)^8 : x = (-3/5)^6
=> (3/5)^8 : x = (3/5)^6
=> x = (3/5)^8 : (3/5)^6
=> x = (3/5)^2
=> x= 9 /25
a) \(3\left(2x-1\right)+1=\left(-2\right)^2-3\left(-2\right)^3\)
\(\Leftrightarrow6x-3+1=4+24\)
\(\Leftrightarrow6x=4+24-1+3\)
\(\Leftrightarrow6x=30\)
\(\Leftrightarrow x=5\)
b) \(\left(x-2\right)\left(x+3\right)>0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2>0\\x+3>0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x>2\\x>-3\end{cases}}\)
c) \(x^2\left(x+2\right)-9\left(x+2\right)=0\)
\(\Leftrightarrow\left(x^2-9\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x^2-9=0\\x+2=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=\pm3\\x=-2\end{cases}}\)
Đăng từng bài thoy nha pn!!!
Bài 1:
Có : 2009 = 2008 + 1 = x + 1
Thay 2009 = x + 1 vào biểu thức trên,ta có :
x\(^5\)- 2009x\(^4\)+ 2009x\(^3\)- 2009x\(^2\)+ 2009x - 2010
= x\(^5\)- (x + 1)x\(^4\)+ (x + 1)x\(^3\)- (x +1)x\(^2\)+ (x + 1) x - (x + 1 + 1)
= x\(^5\)- x\(^5\)- x\(^4\)+ x\(^4\)- x\(^3\)+ x\(^3\)- x\(^2\)+ x\(^2\)+ x - x -1 - 1
= -2
Ta có: f(x) = g(x)
<=> ax3 + 4x(x2 - 1) + 8 = x3 - 4x(bx + 1) + c - 3
<=> ax3 + 4x3 - 4x + 8 = x3 - 4bx2 - 4x + c - 3
<=> (a + 4)x3 - 4x + 8 = x3 - 4bx2 - 4x + c - 3
<=> (a + 4)x3 + 8 = x3 - 4bx2 + c - 3
Đồng nhất hệ số
\(\hept{\begin{cases}a+4=1\\-4b=0\\c-3=8\end{cases}}\) <=> \(\hept{\begin{cases}a=-3\\b=0\\c=11\end{cases}}\)
\(a.\)
\(\left(x-8\right)\left(x^3+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x^3+8=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x^3=-8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
\(S=\left\{8,-2\right\}\)
\(b.\)
\(\left(4x-3\right)-\left(x+5\right)=3\cdot\left(10-x\right)\)
\(\Leftrightarrow4x-3-x-5-30+3x=0\)
\(\Leftrightarrow6x-38=0\)
\(\Leftrightarrow x=\dfrac{38}{6}\)
\(S=\left\{\dfrac{38}{6}\right\}\)
a) \(\left(x-8\right)\left(x^3+8\right)=0\)
=>\(x-8=0 => x=8\)
hoặc \(x^3+8=0\)=>\(x=-2\)
b) \(\left(4x-3\right)-\left(x+5\right)=3\left(10-x\right)\)
\(< =>3x-8=3\left(10-x\right)\)
\(< =>3x-8-30+3x=0\)
\(< =>6x=38=>x=\dfrac{38}{6}=\dfrac{19}{3}\)