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a) Theo bài ra:
c = 1 (1)
a - b = 100 ~> a= 1000+b (2)
Thay (1) và (2) vào A, ta có:
A = 1000+b(b+1) - b(1000+b+1) + 1(1000+b-b)
A = (1000 + b).b + 1000+b - 1000b - \(b^2\) -b + 1000
A= 1000b + \(b^2\) + 1000+b - 1000b - \(b^2 \) - b + 1000
A = (1000b - 1000b) + (\(b^2 - b^2 \))+ (1000 + b - b +1000)
A = 0 + 0 + 0
A = 0
Vậy A = 0
1. Tìm n, biết:
a) \(\dfrac{-32}{\left(-2\right)^n}=4\)
\(\Rightarrow\dfrac{\left(-2\right)^5}{\left(-2\right)^n}=\left(-2\right)^2\)
\(\Rightarrow\left(-2\right)^n.\left(-2\right)^2=\left(-2\right)^5\)
(-2)n + 2 = (-2)5
n + 2 = 5
n = 5 - 2
n = 3.
b) \(\dfrac{8}{2^n}=2\)
\(\Rightarrow\dfrac{2^3}{2^n}=2\)
\(\Rightarrow\) 2n . 2 = 23
n + 1 = 3
n = 3 - 1
n = 2.
c) \(\left(\dfrac{1}{2}\right)^{2n-1}=\dfrac{1}{8}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^{2n-1}=\left(\dfrac{1}{2}\right)^3\)
2n - 1 = 3
2n = 3 + 1
2n = 4
n = 4 : 2
n = 2.
2. Tính:
a) \(\left(\dfrac{1}{2}\right)^3.\left(\dfrac{1}{4}\right)^2\)
\(=\left(\dfrac{1}{2}\right)^3.\left[\left(\dfrac{1}{2}\right)^2\right]^2\)
\(=\left(\dfrac{1}{2}\right)^3.\left(\dfrac{1}{2}\right)^4\)
\(=\left(\dfrac{1}{2}\right)^7\)
\(=\dfrac{1}{128}\)
b) 273 : 93
= (33)3 : (32)3
= 39 : 36
= 33
= 27
c) 1252 : 253
= (53)2 : (52)3
= 56 : 56
= 1
d) \(\dfrac{27^2.8^5}{6^6.32^3}\)
\(=\dfrac{\left(3^3\right)^2.\left(2^3\right)^5}{6^6.\left(2^5\right)^3}\)
\(=\dfrac{3^6.2^{15}}{6^6.2^{15}}\)
\(=\dfrac{3^6}{6^6}\)
\(=\dfrac{1}{64}.\)
B2 :
b) 27\(^3\): 9\(^3\)= (27:9)\(^3\)= 3\(^3\)
c) 125\(^2\): 25\(^3\)= 15625 : 15625 = 1
1. Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\Rightarrow\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\) \(\left(1\right)\)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{b^2.k^2+d^2.k^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\) \(\left(2\right)\)
Từ \(\left(1\right)\text{và (2)}\) \(\Rightarrow\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{ac}{bd}\)
2. \(\left|5-\dfrac{3}{4}x\right|+\left|\dfrac{2}{7}y+3\right|=0\)
\(\left\{{}\begin{matrix}\left|5-\dfrac{3}{4}x\right|\ge0\\\left|\dfrac{2}{7}y+3\right|\ge0\end{matrix}\right.\Rightarrow\left|5-\dfrac{3}{4}x\right|+\left|\dfrac{2}{7}y+3\right|\ge0\)
\(\text{Mà }\left|5-\dfrac{3}{4}x\right|+\left|\dfrac{2}{7}y+3\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|5-\dfrac{3}{4}x\right|=0\\\left|\dfrac{2}{7}y+3\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}5-\dfrac{3}{4}x=0\\\dfrac{2}{7}y+3=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{4}x=5\\\dfrac{2}{7}x=-3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{20}{3}\\y=-\dfrac{21}{2}\end{matrix}\right.\)
\(\text{Vậy }\left\{{}\begin{matrix}x=\dfrac{20}{3}\\y=-\dfrac{21}{2}\end{matrix}\right.\)
3. \(\dfrac{1}{2}a=\dfrac{2}{3}b=\dfrac{3}{4}c\)
\(\Rightarrow\dfrac{a}{2}=\dfrac{b}{\dfrac{3}{2}}=\dfrac{c}{\dfrac{4}{3}}\)
\(\text{Mà }a-b=15\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{2}=\dfrac{b}{\dfrac{3}{2}}=\dfrac{c}{\dfrac{4}{3}}=\dfrac{a-b}{2-\dfrac{3}{2}}=\dfrac{15}{\dfrac{1}{2}}=30\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{2}=30\Rightarrow a=30.2=60\\\dfrac{b}{\dfrac{3}{2}}=30\Rightarrow b=30.\dfrac{3}{2}=45\\\dfrac{c}{\dfrac{4}{3}}=30\Rightarrow c=30.\dfrac{4}{3}=40\end{matrix}\right.\)
\(\text{Vậy }\left\{{}\begin{matrix}a=60\\b=45\\c=40\end{matrix}\right.\)
Lời giải:
$a^2+ab=c^2+bc$
$\Rightarrow a(a+b)=c(b+c)\Rightarrow \frac{a+b}{c}=\frac{b+c}{a}(1)$
$a^2+ac=b^2+bc$
$\Rightarrow a(a+c)=b(b+c)\Rightarrow \frac{a+c}{b}=\frac{b+c}{a}(2)$
Từ $(1); (2)\Rightarrow \frac{a+b}{c}=\frac{b+c}{a}+\frac{c+a}{b}$
Áp dụng TCDTSBN:
$\frac{a+b}{c}=\frac{b+c}{a}+\frac{c+a}{b}=\frac{a+b+b+c+c+a}{c+a+b}=\frac{2(a+b+c)}{a+b+c}=2$
$\Rightarrow a+b=2c; b+c=2a; c+a=2b$
$\Rightarrow a+b-(b+c)=2c-2a$
$\Rightarrow a-c=2c-2a\Rightarrow 3a=3c\Rightarrow a=c$
$2b=c+a=a+a=2a\Rightarrow a=b$
Vậy $a=b=c$
Do đó:
$K=(1+\frac{a}{a})(1+\frac{a}{a})(1+\frac{a}{a})=(1+1)(1+1)(1+1)=8$