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\(B=\dfrac{3}{1x2}+\dfrac{3}{2x3}+...+\dfrac{3}{50x51}\)
\(B=3x\left(\dfrac{1}{1x2}+\dfrac{1}{2x3}+...+\dfrac{1}{50x51}\right)\)
\(B=3x\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{50}-\dfrac{1}{51}\right)\)
\(B=3x\left(1-\dfrac{1}{51}\right)=3x\dfrac{50}{51}=\dfrac{150}{51}\)
đọc lại kĩ đi làm bài 1,2,3 mà làm chắc bài ba
Bài 11:
a: \(A=\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\frac{\sqrt{a}-\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)-\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{a-1-a+4}=\frac{\sqrt{a}-2}{3\sqrt{a}}\)
b: A>0
=>\(\frac{\sqrt{a}-2}{3\sqrt{a}}>0\)
=>\(\sqrt{a}-2>0\)
=>\(\sqrt{a}>2\)
=>a>4
c: Đặt B=1:A
\(=1:\frac{\sqrt{a}-2}{3\sqrt{a}}=\frac{3\sqrt{a}}{\sqrt{a}-2}\)
Để B là số nguyên thì \(3\sqrt{a}\) ⋮\(\sqrt{a}-2\)
=>\(3\sqrt{a}-6+6\) ⋮\(\sqrt{a}-2\)
=>6⋮\(\sqrt{a}-2\)
=>\(\sqrt{a}-2\in\left\lbrace1;-1;2;-2;3;-3;6;-6\right\rbrace\)
=>\(\sqrt{a}\) ∈{3;1;4;0;5;-1;8;-4}
=>\(\sqrt{a}\) ∈{0;1;3;4;5;8}
=>a∈{0;1;9;16;25;64}
Kết hợp ĐKXĐ, ta được: a∈{9;16;25;64}
Bài 10:
a: \(A=\frac{\sqrt{x}-3}{\sqrt{x}+3}+\frac{\sqrt{x}+3}{\sqrt{x}-3}-\frac{12\sqrt{x}}{x-9}\)
\(=\frac{\left(\sqrt{x}-3\right)^2+\left(\sqrt{x}+3\right)^2-12\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{x-6\sqrt{x}+9+x+6\sqrt{x}+9-12\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{2x-12\sqrt{x}+18}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{2\left(\sqrt{x}-3\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+3}\)
b: A=1
=>\(2\left(\sqrt{x}-3\right)=\sqrt{x}+3\)
=>\(2\sqrt{x}-6=\sqrt{x}+3\)
=>\(\sqrt{x}=9\)
=>x=81(nhận)
d: Để A là số nguyên thì \(2\left(\sqrt{x}-3\right)\vdots\sqrt{x}+3\)
=>\(2\sqrt{x}-6\vdots\sqrt{x}+3\)
=>\(2\sqrt{x}+6-12\vdots\sqrt{x}+3\)
=>\(-12\vdots\sqrt{x}+3\)
=>\(\sqrt{x}+3\in\left\lbrace3;4;6;12\right\rbrace\)
=>\(\sqrt{x}\in\left\lbrace0;1;3;9\right\rbrace\)
=>x∈{0;1;9;81}
Kết hợp ĐKXĐ, ta được: x∈{0;1;81}
a) \(\dfrac{3}{5}+x=\dfrac{5}{4}+\dfrac{3}{4}\)
\(\dfrac{3}{5}+x=2\)
\(x=2-\dfrac{3}{5}\)
\(x=\dfrac{7}{5}\)
b) \(x-\dfrac{3}{2}+\dfrac{5}{4}=\dfrac{9}{2}\)
\(x-\dfrac{3}{2}=\dfrac{9}{2}-\dfrac{5}{4}\)
\(x-\dfrac{3}{2}=\dfrac{13}{4}\)
\(x=\dfrac{13}{4}+\dfrac{3}{2}\)
\(x=\dfrac{19}{4}\)
c) \(x\div\dfrac{9}{7}=\dfrac{3}{5}\)
\(x=\dfrac{3}{5}\times\dfrac{9}{7}\)
\(x=\dfrac{27}{35}\)