Tìm a;b
b) 2^a + 342 = 7^b
c) 2^a + 80 = 3^b
d) 5^a + 9999 = 20b
e) 10^a + 168 = b^2
f) 5^a + 323 = b^2
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ĐKXĐ: \(x\ge0;x\ne4\)
\(A=\dfrac{x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
b. \(x=36\Rightarrow A=\dfrac{\sqrt{36}}{\sqrt{36}-2}=\dfrac{6}{6-2}=\dfrac{3}{2}\)
c. \(A=-\dfrac{1}{3}\Rightarrow\dfrac{\sqrt{x}}{\sqrt{x}-2}=-\dfrac{1}{3}\Rightarrow3\sqrt{x}=2-\sqrt{x}\)
\(\Rightarrow4\sqrt{x}=2\Rightarrow\sqrt{x}=\dfrac{1}{2}\Rightarrow x=\dfrac{1}{4}\)
d. \(A>0\Rightarrow\dfrac{\sqrt{x}}{\sqrt{x}-2}>0\Rightarrow\sqrt{x}-2>0\Rightarrow x>4\)
e. \(A=\dfrac{\sqrt{x}-2+2}{\sqrt{x}-2}=1+\dfrac{2}{\sqrt{x}-2}\in Z\Rightarrow\sqrt{x}-2=Ư\left(2\right)\)
\(\Rightarrow\sqrt{x}-2=\left\{-2;-1;1;2\right\}\)
\(\Rightarrow\sqrt{x}=\left\{0;1;3;4\right\}\Rightarrow x=\left\{0;1;9;16\right\}\)
a: Ta có: \(A=\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)
\(=\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
b: Thay x=36 vào A, ta được:
\(A=\dfrac{6}{6-2}=\dfrac{6}{4}=\dfrac{3}{2}\)
c: Để \(A=-\dfrac{1}{3}\) thì \(3\sqrt{x}=-\sqrt{x}+2\)
\(\Leftrightarrow4\sqrt{x}=2\)
hay \(x=\dfrac{1}{4}\)
a: Để A là phân số thì n+5<>0
hay n<>-5
b: Để A=-1/2 thì n-1/n+5=-1/2
=>2n-2=-n-5
=>3n=-3
hay n=-1
c: Để A là số nguyên thì \(n-1⋮n+5\)
\(\Leftrightarrow n+5\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
hay \(n\in\left\{-4;-6;-3;-7;-2;-8;1;-11\right\}\)
a: A=3,5:40%=8,75
b: 1/4 của A là 1/4*8,75=2,1875
c: 75% của A là 3/4*8,75=6,5625
Ta có: \(A=\frac{6n-9+13}{2n-3}=\frac{3\left(2n-3\right)+13}{2n-3}\)
Mà: 3 ( 2n - 3 ) chia hết cho 2n - 3
=> 13 chia hết cho 2n - 3 => 2n - 3 E Ư(13) = {1,-1,13,-13}
=> 2n E {4,2,16,-10}
Ta có bảng sau:
2n | 4 | 2 | 16 | -10 |
n | 2 | 1 | 8 | -5 |
Tớ nghĩ là cộng vì dấu ''+'' nằm dưới dấu ''='' mà, chắc là quên ấn nút ''Shift'' ấy mà!
abc:(a+b+c)=100
aba=(a+b+c)x100
abc=a x100+bx100+cx100
ax100+bx10+c=ax100+bx100+cx100
( đề có vẻ sai )
abc:(a+b+c)=100
aba=(a+b+c)x100
abc=a x100+bx100+cx100
ax100+bx10+c=ax100+bx100+cx100
( đề có vẻ sai ) Nếu bn cảm thấy đúng thì k cho mình nhé!Học Tốt
b) Ta có:
\(7^b=2^a+342\)
\(\Rightarrow\left\{{}\begin{matrix}7^b=343\\2^a=7^b-342\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}7^b=7^3\\2^a=343-342\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}b=3\\2^a=2^0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}b=3\\a=0\end{matrix}\right.\)
c) Ta có:
\(2^a+80=3^b\)
\(\Rightarrow\left\{{}\begin{matrix}3^b=81\\2^a=3^b-80\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}3^b=3^4\\2^a=81-80\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}b=4\\2^a=2^0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}b=4\\a=0\end{matrix}\right.\)
d) Ta có:
\(5^a+9999=20b\)
\(\Rightarrow\left\{{}\begin{matrix}5^a=1\\20b=9999+5^a\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}5^a=5^0\\20b=9999+1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=0\\b=\dfrac{10000}{20}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=0\\b=500\end{matrix}\right.\)
e) \(10^a+168=b^2\)
\(\Rightarrow\left\{{}\begin{matrix}10^a=1\\b^2=168+10^a\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}10^a=10^0\\b^2=168+1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=0\\b^2=169\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=0\\\left[{}\begin{matrix}b=13\\b=-13\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left(a;b\right)=\left(0;13\right);\left(0;-13\right)\)
f) \(5^a+323=b^2\)
\(\Rightarrow\left\{{}\begin{matrix}5^a=1\\b^2=5^a+323\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}5^a=5^0\\b^2=324\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=0\\b^2=18^2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=0\\\left[{}\begin{matrix}b=18\\b=-18\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left(a;b\right)=\left(0;18\right);\left(0;-18\right)\)
b) a = 0
b = 3
c) a = 0
b = 4
d) a = 0
b = 500
e) a = 0
b ∈ {13; -13}
f) a = 0
b ∈ {18; -18}