Ai giúp t câu này vs

Ta có \(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=a+b+c+2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\Leftrightarrow7^2=23+2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\Leftrightarrow\sqrt{ab}+\sqrt{bc}+\sqrt{ca}=13\)

Ta lại có \(\sqrt{a}+\sqrt{b}+\sqrt{c}=7\Leftrightarrow\sqrt{c}-6=-\sqrt{a}-\sqrt{b}+1\Leftrightarrow\sqrt{ab}+\sqrt{c}-6=\sqrt{ab}-\sqrt{a}-\sqrt{b}+1=\sqrt{a}\left(\sqrt{b}-1\right)-\left(\sqrt{b}-1\right)=\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)\)

Chứng minh tương tự:

\(\sqrt{bc}+\sqrt{a}-6=\left(\sqrt{b}-1\right)\left(\sqrt{c}-1\right)\)

\(\sqrt{ac}+\sqrt{b}-6=\left(\sqrt{a}-1\right)\left(\sqrt{c}-1\right)\)

Vậy A=\(\dfrac{1}{\sqrt{ab}+\sqrt{c}-6}+\dfrac{1}{\sqrt{bc}+\sqrt{a}-6}+\dfrac{1}{\sqrt{ca}+\sqrt{b}-6}=\dfrac{1}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)}+\dfrac{1}{\left(\sqrt{b}-1\right)\left(\sqrt{c}-1\right)}+\dfrac{1}{\left(\sqrt{c}-1\right)\left(\sqrt{a}-1\right)}=\dfrac{\sqrt{c}-1+\sqrt{a}-1+\sqrt{b}-1}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)\left(\sqrt{c}-1\right)}=\dfrac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)-3}{\sqrt{abc}+\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)-\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)}=\dfrac{7-3}{3+7-13-1}=-1\)

Câu hỏi của hoàng thị huyền trang - Toán lớp 9 - Học toán với OnlineMath

Em tham khảo nhé!

- Theo BĐT Cauchy ta có:

\(\sqrt{a.1}\le\dfrac{a+1}{2}\)

\(\sqrt{b.1}\le\dfrac{b+1}{2}\)

\(\sqrt{c.1}\le\dfrac{c+1}{2}\)

\(\sqrt{ab}\le\dfrac{a+b}{2}\)

\(\sqrt{bc}\le\dfrac{b+c}{2}\)

\(\sqrt{ca}\le\dfrac{c+a}{2}\)

\(\Rightarrow\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\le\dfrac{3\left(a+b+c\right)+3}{2}=\dfrac{3.3+3}{2}=6\)

Dấu "=" xảy ra khi \(a=b=c=1\)

Mà ta có: \(\sqrt{a}+\sqrt{b}+\sqrt{c}+\sqrt{ab}+\sqrt{bc}+\sqrt{ca}=6\)

\(\Rightarrow a=b=c=1\)

\(M=\dfrac{a^{30}+b^4+c^{1975}}{a^{30}+b^4+c^{2023}}=\dfrac{1^{30}+1^4+1^{1975}}{1^{30}+1^4+1^{2023}}=1\)

Từ giả thiết: \(\sqrt{a}+\sqrt{b}+\sqrt{c}=7\Leftrightarrow\sqrt{c}=7-\sqrt{a}-\sqrt{b}\)

Xét hạng tử: \(\frac{1}{\sqrt{ab}+\sqrt{c}-6}=\frac{1}{\sqrt{ab}+7-\sqrt{a}-\sqrt{b}-6}=\frac{1}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)}\)

Từ đó: \(N=\frac{1}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)}+\frac{1}{\left(\sqrt{b}-1\right)\left(\sqrt{c}-1\right)}+\frac{1}{\left(\sqrt{c}-1\right)\left(\sqrt{a}-1\right)}\)

\(=\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}-3}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)\left(\sqrt{c}-1\right)}=\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}-3}{\sqrt{abc}-\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)+\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)-1}\)

\(=\frac{7-3}{3-\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)+7-1}=\frac{4}{9-\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)}\)

Mặt khác: \(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}=\frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2-\left(a+b+c\right)}{2}=13\)

Suy ra: \(N=\frac{4}{9-13}=-1\). Kết luận: N = -1.

Từ giả thiết: \sqrt{a}+\sqrt{b}+\sqrt{c}=7\Leftrightarrow\sqrt{c}=7-\sqrt{a}-\sqrt{b}a​+b​+c​=7⇔c​=7−a​−b​

Xét hạng tử: \frac{1}{\sqrt{ab}+\sqrt{c}-6}=\frac{1}{\sqrt{ab}+7-\sqrt{a}-\sqrt{b}-6}=\frac{1}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)}ab​+c​−61​=ab​+7−a​−b​−61​=(a​−1)(b​−1)1​

Từ đó: N=\frac{1}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)}+\frac{1}{\left(\sqrt{b}-1\right)\left(\sqrt{c}-1\right)}+\frac{1}{\left(\sqrt{c}-1\right)\left(\sqrt{a}-1\right)}N=(a​−1)(b​−1)1​+(b​−1)(c​−1)1​+(c​−1)(a​−1)1​

=\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}-3}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)\left(\sqrt{c}-1\right)}=\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}-3}{\sqrt{abc}-\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)+\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)-1}=(a​−1)(b​−1)(c​−1)a​+b​+c​−3​=abc​−(ab​+bc​+ca​)+(a​+b​+c​)−1a​+b​+c​−3​

=\frac{7-3}{3-\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)+7-1}=\frac{4}{9-\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)}=3−(ab​+bc​+ca​)+7−17−3​=9−(ab​+bc​+ca​)4​

Mặt khác: \sqrt{ab}+\sqrt{bc}+\sqrt{ca}=\frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2-\left(a+b+c\right)}{2}=13ab​+bc​+ca​=2(a​+b​+c​)2−(a+b+c)​=13

Suy ra: N=\frac{4}{9-13}=-1N=9−134​=−1. Kết luận: N = -1.

Ta có \(\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2=a+b+c+2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\)

mà \(\sqrt{a}+\sqrt{b}+\sqrt{c}=7;a+b+c=23\)nên \(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}=13\)

Ta có: \(\sqrt{a}+\sqrt{b}+\sqrt{c}=7\Rightarrow\sqrt{c}-6=-\sqrt{a}-\sqrt{b}+1\)

nên \(\sqrt{ab}+\sqrt{c}-6=\sqrt{ab}-\sqrt{a}-\sqrt{b}+1=\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)\)

Tương tự \(\hept{\begin{cases}\sqrt{bc}+\sqrt{a}-6=\left(\sqrt{b}-1\right)\left(\sqrt{c}-1\right)\\\sqrt{ac}+\sqrt{b}-6=\left(\sqrt{a}-1\right)\left(\sqrt{c}-1\right)\end{cases}}\)

Vậy \(H=\frac{1}{\sqrt{ab}+\sqrt{c}-6}+\frac{1}{\sqrt{bc}+\sqrt{a}-6}+\frac{1}{\sqrt{ac}+\sqrt{b}-6}\)

\(=\frac{1}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)}+\frac{1}{\left(\sqrt{b}-1\right)\left(\sqrt{c}-1\right)}+\frac{1}{\left(\sqrt{c}-1\right)\left(\sqrt{a}-1\right)}\)

\(=\frac{\sqrt{c}-1+\sqrt{a}-1+\sqrt{b}-1}{\left(\sqrt{a}-1\right)\left(\sqrt{b}-1\right)\left(\sqrt{c}-1\right)}\)

\(=\frac{\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)-3}{\sqrt{abc}+\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)-\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)}=\frac{7-3}{3+7-13-1}=-1\)

Ta có : \(b=\dfrac{c+a}{2}\Rightarrow2b=c+a\Rightarrow a-b=b-c\)

Dó đó : \(P=\left(\dfrac{1}{\sqrt{a}+\sqrt{b}}+\dfrac{1}{\sqrt{b}+\sqrt{c}}\right)\left(\sqrt{a}+\sqrt{c}\right)\)

\(P=\left[\dfrac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}+\dfrac{\sqrt{b}-\sqrt{c}}{\left(\sqrt{b}+\sqrt{c}\right)\left(\sqrt{b}-\sqrt{c}\right)}\right]\left(\sqrt{a}+\sqrt{c}\right)\)

\(P=\left[\dfrac{\sqrt{a}-\sqrt{b}}{a-b}+\dfrac{\sqrt{b}-\sqrt{c}}{b-c}\right]\left(\sqrt{a}+\sqrt{c}\right)\)

\(P=\left[\dfrac{\sqrt{a}-\sqrt{b}}{b-c}+\dfrac{\sqrt{b}-\sqrt{c}}{b-c}\right]\left(\sqrt{a}+\sqrt{c}\right)\) Vì  \(\left(a-b=b-c\right)\)

\(P=\left[\dfrac{\sqrt{a}-\sqrt{b}+\sqrt{b}-\sqrt{c}}{b-c}\right]\left(\sqrt{a}+\sqrt{c}\right)\)

\(P=\dfrac{\sqrt{a}-\sqrt{c}}{b-c}\left(\sqrt{a}+\sqrt{c}\right)\)

\(P=\dfrac{a-c}{a-b}=\dfrac{a-c}{a-\dfrac{a+c}{2}}=\dfrac{a-c}{\dfrac{2a-a-c}{2}}=\dfrac{a-c}{\dfrac{a-c}{2}}=2\)

\(\sqrt{\dfrac{ab}{c+ab}}=\sqrt{\dfrac{ab}{c\left(a+b+c\right)+ab}}=\sqrt{\dfrac{ab}{\left(a+c\right)\left(b+c\right)}}\le\dfrac{1}{2}\left(\dfrac{a}{a+c}+\dfrac{b}{b+c}\right)\)

Tương tự: \(\sqrt{\dfrac{bc}{a+bc}}\le\dfrac{1}{2}\left(\dfrac{b}{a+b}+\dfrac{c}{a+c}\right)\) ; \(\sqrt{\dfrac{ca}{b+ca}}\le\dfrac{1}{2}\left(\dfrac{a}{a+b}+\dfrac{c}{b+c}\right)\)

Cộng vế với vế:

\(P\le\dfrac{1}{2}\left(\dfrac{a}{a+c}+\dfrac{c}{a+c}+\dfrac{b}{b+c}+\dfrac{c}{b+c}+\dfrac{b}{a+b}+\dfrac{a}{a+b}\right)=\dfrac{3}{2}\)

Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)

b) Thay x=49 vào A, ta được:

\(A=\dfrac{7-1}{7-5}=\dfrac{6}{2}=3\)

a) Ta có: \(B=\dfrac{\sqrt{x}+3}{\sqrt{x}+1}+\dfrac{5}{\sqrt{x}-1}+\dfrac{4}{x-1}\)

\(=\dfrac{x+2\sqrt{x}-3+5\sqrt{x}+5+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{x+7\sqrt{x}+6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\sqrt{x}+6}{\sqrt{x}-1}\)