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\(a,\Leftrightarrow x=7-4=3\\ b,\Leftrightarrow2x=-18+5=-13\\ \Leftrightarrow x=-\dfrac{13}{2}\\ c,\Leftrightarrow x-21=10\\ \Leftrightarrow x=31\\ d,\Leftrightarrow-12-x+19=0\\ \Leftrightarrow7-x=0\\ \Leftrightarrow x=7\)
a, <=> x=7-4
<=> x=3
b, 2x= -18 +5
<=>2x=-13
<=> x= -13/2
c, <=> x -21=-10
<=> x= -10 +21
<=> x=11
d, <=> -12+19 -x=0
<=> 7-x=0
<=> x=7
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a) \(a^2\cdot a^3\cdot a^7\cdot b^2\cdot b\)
\(=\left(a^2\cdot a^3\cdot a^7\right)\cdot\left(b^2\cdot b\right)\)
\(=a^{12}\cdot b^3\)
b) \(b^6\cdot b\cdot c^7\cdot c^8\)
\(=\left(b^6\cdot b\right)\cdot\left(c^7\cdot c^8\right)\)
\(=b^7\cdot c^{15}\)
c) \(a^8\cdot a^9\cdot a\cdot c\cdot c^{20}\)
\(=\left(a^8\cdot a^9\cdot a\right)\cdot\left(c\cdot c^{20}\right)\)
\(=a^{18}\cdot c^{21}\)
d) \(a^2\cdot a^3\cdot b^4\cdot c\cdot c^3\)
\(=\left(a^2\cdot a^3\right)\cdot b^4\cdot\left(c\cdot c^3\right)\)
\(=a^5\cdot b^4\cdot c^4\)
a) Kiểm tra lại nhé
b) \(b^6.b^7.c^8\)
\(=b^{6+7}.c^8=b^{13}.c^8\)
c) \(a^8.a^9.a.c.c^{20}\)
\(=a^{8+9+1}.c^{1+20}\)
\(=a^{18}.c^{21}\)
d) \(a^2.a^3.b^4.c.c^3\)
\(=a^{2+3}.b^4.c^{1+3}\)
\(=a^5.b^4.c^4\)
\(#WendyDang\)
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Vì ( x + 23 ) . ( x - 8 ) = 0
nên ( x + 23 ) hoặc ( x - 8 ) bằng 0
x + 23 = 0 x - 8 = 0
x = 0 - 23 x = 0 + 8
x = -23 x = 8
Vậy x thuộc { -23;8}
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\((3x-2)-2^5\times9=7^2\\\Rightarrow (3x-2)-32\times9=49\\\Rightarrow (3x-2)-288=49\\\Rightarrow 3x-2=49+288\\\Rightarrow 3x-2=337\\\Rightarrow 3x=337+2\\\Rightarrow 3x=339\\\Rightarrow x=\dfrac{339}{3}\\\Rightarrow x=113\)
#\(Toru\)
(3.x - 2) - 2⁵ × 9=7²
=>(3.x-2)-32 x 9=49
=>(3.x-2)-288=49
=>3.x-2=337
=>3.x=339
=>x=113
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a) - lx - 60l - 61 = 0
-61 - 0 = lx - 60l
-61 = lx - 60l mà lx - 60l\(\ge\)0 với mọi x .
\(\Rightarrow x\in\varnothing\)
Vậy\(x\in\varnothing\)
X=0+7/9
X=7/9