giải giúp tui!!Please!!

Cai nay lop 7 ma, Ap dung cong thuc tinh canh huyen tam giac vuong la ra duong cheo cua cac tam giac trong hinh thang

\(x^2\sqrt{9-x^2}\)

Ta có:

\(9-x^2\le9\forall x\)

\(Minx^2\sqrt{9-x^2}=0\Leftrightarrow x=0\)

- Gọi O là giao điểm của AC và BD.

- AB//CD nên góc BAC = góc ACD (so le trong), tương tự góc ABD=góc BDC.

- Theo đề bài góc ACD=gócBDC nên góc BAC=góc ABD.

=>Tam giác ABO cân tại O

=> 0A=0B.(1)

Tương tự tam giác ODC cân tại O

=>OD=OC.(2)

Lại có góc AOD=góc BOC (đối đỉnh )

(3) Từ (1), (2), (3) suy ra tam giác AOD = tam giác OBC

nên suy ra :

=> AD = BC 

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

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

\(=-x-1\)

Với `a > 0,a \ne 1` có:

    `(1/[a-\sqrt{a}]+1/[\sqrt{a}-1]):[\sqrt{a}+1]/[a-2\sqrt{a}+1]`

`=[1+\sqrt{a}]/[\sqrt{a}(\sqrt{a}-1)].[(\sqrt{a}-1)^2]/[\sqrt{a}+1]`

`=[\sqrt{a}-1]/\sqrt{a}`

\(=\left(\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\dfrac{1}{\sqrt{a}-1}\right)\times\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\)

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

\(=\dfrac{\sqrt{a}-1}{1}\)

Đk: \(a\ge0;a\ne1\)

\(\left(\dfrac{1}{a-\sqrt{a}}+\dfrac{1}{\sqrt{a}-1}\right):\dfrac{\sqrt{a}+1}{a-2\sqrt{a}+1}\\ =\left(\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\dfrac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)^2}\\ =\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)^2}\\ =\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)^2}{\sqrt{a}\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\\ =\dfrac{\sqrt{a}-1}{\sqrt{a}}\)

a) Điều kiện: \(a\ge0;a\ne1\)

\(P=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\\ =\left(\dfrac{\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{\left(1+\sqrt{a}\right)\left(a-\sqrt{a}+1\right)}{1+\sqrt{a}}-\sqrt{a}\right)\\ =\left(a+2\sqrt{a}+1\right)\left(a-2\sqrt{a}+1\right)\\ =\left(\sqrt{a}+1\right)^2\left(\sqrt{a}-1\right)^2\\ =\left(a-1\right)^2\)

b) Để \(P< 7-4\sqrt{3}\Rightarrow P< \left(2-\sqrt{3}\right)^2\Rightarrow\left(a-1\right)^2< \left(2-\sqrt{3}\right)^2\)

\(\Rightarrow-\left(2-\sqrt{3}\right)< a-1< 2-\sqrt{3}\)

\(\Rightarrow\sqrt{3}-1< a< 3-\sqrt{3}\) (Thỏa mãn)

\(B=\sqrt{49x^2-22x+9}+\sqrt{49x^2+22x+9}\)

\(=\sqrt{\left[\left(7x\right)^2-2.7x.\dfrac{11}{7}+\dfrac{121}{49}\right]+\dfrac{320}{49}}+\sqrt{\left[\left(7x\right)^2+2.7x.\dfrac{11}{7}+\dfrac{121}{49}\right]+\dfrac{320}{49}}\)

\(=\)\(\sqrt{\left(\dfrac{11}{7}-7x\right)^2+\left(\dfrac{8\sqrt{5}}{7}\right)^2}+\sqrt{\left(7x+\dfrac{11}{7}\right)^2+\left(\dfrac{8\sqrt{5}}{7}\right)^2}\)(1)

Áp dụng BĐT Mincopxki, ta có:

\(\left(1\right)\ge\sqrt{\left(\dfrac{11}{7}-7x+7x+\dfrac{11}{7}\right)^2+\left(\dfrac{2.8\sqrt{5}}{7}\right)^2}\)

\(=\sqrt{36}=6\)

\(MinB=6\Leftrightarrow...\)

Ta có : \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{2}{\sqrt{xy}}\) (bđt Cauchy) (1) 

Tương tự được \(\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{2}{\sqrt{yz}}\left(2\right);\dfrac{1}{z}+\dfrac{1}{x}\ge\dfrac{2}{\sqrt{xz}}\left(3\right)\)

Cộng (1) ; (2) ; (3) theo vế được 

\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{z}+\dfrac{1}{x}\ge\dfrac{2}{\sqrt{xy}}+\dfrac{2}{\sqrt{yz}}+\dfrac{2}{\sqrt{zx}}\)

\(\Leftrightarrow\dfrac{2}{x}+\dfrac{2}{y}+\dfrac{2}{z}\ge\dfrac{2}{\sqrt{xy}}+\dfrac{2}{\sqrt{yz}}+\dfrac{2}{\sqrt{zx}}\) <=> ĐPCM

"=" khi x = y = z 

Áp dụng BĐT Cô si cho 2 số dương, ta có:

\(\dfrac{1}{x}+\dfrac{1}{y}\ge2.\sqrt{\dfrac{1}{x}.\dfrac{1}{y}}=\dfrac{2}{\sqrt{xy}}\)

Tương tự:

\(\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{2}{\sqrt{yz}}\)

\(\dfrac{1}{z}+\dfrac{1}{x}\ge\dfrac{2}{\sqrt{zx}}\)

\(\Rightarrow2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge2\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\right)\)

\(\Rightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=z\)