\(\left\{{}\begin{matrix}a=\dfrac{35}{49}=\dfrac{5}{7}\\b=\sqrt{\dfrac{5^2}{7^2}}=\dfrac{5}{7}\\c=\dfrac{\sqrt{5^2}+\sqrt{35^2}}{\sqrt{7^2}+\sqrt{49^2}}=\dfrac{5+35}{7+49}=\dfrac{5}{7}\\d=\dfrac{\sqrt{5^2}-\sqrt{35^2}}{\sqrt{7^2}-\sqrt{49^2}}=\dfrac{5-35}{7-49}=\dfrac{5}{7}\end{matrix}\right.\)

\(\Rightarrow a=b=c=d=\dfrac{5}{7}\)

\(a=\dfrac{35}{49};b=\dfrac{5}{7}\\ c,=\dfrac{5+35}{7+49}=\dfrac{12}{14}=\dfrac{6}{7}\\ d,=\dfrac{5-35}{7-49}\)

Áp dụng t/c dtsbn:

\(\dfrac{5}{7}=\dfrac{35}{49}=\dfrac{5+35}{7+49}=\dfrac{5-35}{7-49}\) hay \(a=b=c=d\)

đoạn cuối thiếu dấu"+"

\(A=\dfrac{\sqrt{4}-\sqrt{5}}{4-5}+\dfrac{\sqrt{5}-\sqrt{6}}{5-6}+....+\dfrac{\sqrt{34}-\sqrt{35}}{34-35}+\dfrac{\sqrt{35}-\sqrt{36}}{335-36}\)

\(A=\dfrac{\sqrt{4}-\sqrt{5}+\sqrt{5}-\sqrt{6}+....+\sqrt{35}-\sqrt{36}}{-1}=\dfrac{\sqrt{4}-\sqrt{36}}{-1}\)

\(A=\sqrt{36}-\sqrt{4}=6-2=4\)

\(\dfrac{1}{\sqrt{4}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{6}}+\dfrac{1}{\sqrt{6}+\sqrt{7}}+...+\dfrac{1}{\sqrt{34}+\sqrt{35}}+\dfrac{1}{\sqrt{35}+\sqrt{36}}\)

\(=-\sqrt{4}+\sqrt{5}-\sqrt{5}+\sqrt{6}-...-\sqrt{35}+\sqrt{36}\)

\(=6-2=4\)

Bài 1:

a) Ta có: \(\left(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{2}\sqrt{20}-\dfrac{5}{4}\sqrt{\dfrac{4}{5}}+\sqrt{5}\right)\)

\(=\left(\sqrt{5}+\sqrt{5}-\dfrac{5}{4}\cdot\dfrac{2}{\sqrt{5}}+\sqrt{5}\right)\)

\(=3\sqrt{5}-\dfrac{1}{2}\sqrt{5}\)

\(=\dfrac{5}{2}\sqrt{5}\)

c) Ta có: \(\dfrac{5\sqrt{7}-7\sqrt{5}+2\sqrt{70}}{\sqrt{35}}\)

\(=\dfrac{\sqrt{35}\left(\sqrt{5}-\sqrt{7}+2\sqrt{2}\right)}{\sqrt{35}}\)

\(=2\sqrt{2}+\sqrt{5}-\sqrt{7}\)

Bài 2:

e) ĐKXĐ: \(\dfrac{4}{3}\le x\le6\)

Ta có: \(\sqrt{6-x}=3x-4\)

\(\Leftrightarrow6-x=\left(3x-4\right)^2\)

\(\Leftrightarrow9x^2-24x+16+6-x=0\)

\(\Leftrightarrow9x^2-25x+22=0\)

\(\Delta=\left(-25\right)^2-4\cdot9\cdot22=625-792< 0\)

Vậy: Phương trình vô nghiệm

a: \(A=\left(2\sqrt{5}-3\sqrt{5}+3\sqrt{5}\right)\cdot\sqrt{5}=2\sqrt{5}\cdot\sqrt{5}=10\)

\(B=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\)

\(=\sqrt{x}-1+\sqrt{x}=2\sqrt{x}-1\)

b: A=2B

=>\(10=4\sqrt{x}-2\)

=>\(4\sqrt{x}=12\)

=>x=9(nhận)

\(a.A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}=\dfrac{2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}=\dfrac{2}{x+\sqrt{x}+1}\left(x\ge0;x\ne1\right)\)

Để : \(A=\dfrac{2}{7}\Leftrightarrow\dfrac{2}{x+\sqrt{x}+1}=\dfrac{2}{7}\)

\(\Leftrightarrow x+\sqrt{x}-6=0\)

\(\Leftrightarrow x-2\sqrt{x}+3\sqrt{x}-6=0\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)=0\)

\(\Leftrightarrow x=4\left(TM\right)\)

\(b.A^2=\left(\dfrac{2}{x+\sqrt{x}+1}\right)^2=\dfrac{4}{\left(x+\sqrt{x}+1\right)^2}\left(1\right)\)

\(2A=2.\dfrac{2}{x+\sqrt{x}+1}=\dfrac{4}{x+\sqrt{x}+1}\left(2\right)\)

Mà : \(x+\sqrt{x}+1\le\left(x+\sqrt{x}+1\right)^2\left(3\right)\)

Từ \(\left(1;2;3\right)\Rightarrow2A\ge A^2\)

Bài 3: Gọi số học sinh giỏi,khá,trung bình lần lượt là a,b,c

Theo bài ra ta có : \(\dfrac{a}{b}=\dfrac{2}{3}\Rightarrow\dfrac{a}{2}=\dfrac{b}{3}\); \(\dfrac{b}{c}=\dfrac{4}{5}\Rightarrow\dfrac{b}{4}=\dfrac{c}{5}\)

\(\Rightarrow\dfrac{a}{2}=\dfrac{b}{3};\dfrac{b}{4}=\dfrac{c}{5}\)

\(\Rightarrow\dfrac{a}{8}=\dfrac{b}{12}=\dfrac{c}{15}\); \(a+b+c=35\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

\(\dfrac{a}{8}=\dfrac{b}{12}=\dfrac{c}{15}=\dfrac{a+b+c}{8+12+15}=\dfrac{35}{35}=1\)

Ta có : \(\dfrac{a}{8}=1\Rightarrow a=8\)

Làm tương tự ta tính được : \(b=12;c=15\)

Vậy số học sinh giỏi là 8 bạn

Số học sinh khá là 12 bạn

Số học sinh trung bình là 15 bạn

Bài 1:

\(\sqrt{1}-\sqrt{4}+\sqrt{9}-\sqrt{16}+\sqrt{25}-\sqrt{36}+.....-\sqrt{400}\)

\(=1-2+3-4+5-6+.....-20\)

\(=\left(1-2\right)+\left(3-4\right)-\left(5-6\right)+.....+\left(19-20\right)\)

\(=\left(-1\right)\times\dfrac{\dfrac{\left(20-1\right)\times1+1}{2}}{2}\)

\(=\left(-1\right)\times10\)

\(=-10\)

Dễ thế này mà ko ai lm à

Chúc bn học tốt

a) 2√6>3√2>√13>2√326

b)1/3√39>1/4√32>1/5√35>1/2√51339

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