a) \(P=\dfrac{1}{\sqrt{5}-2}+\dfrac{1}{\sqrt{5}+2}=\dfrac{\sqrt{5}+2+\sqrt{5}-2}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}=\dfrac{2\sqrt{5}}{\left(\sqrt{5}\right)^2-2^2}=2\sqrt{5}\)

b)\(Q=\left(1+\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right)\cdot\dfrac{1}{\sqrt{x}}=\dfrac{\sqrt{x}-1+\sqrt{x}+1}{\sqrt{x}-1}\cdot\dfrac{1}{\sqrt{x}}\)

\(Q=\dfrac{2\sqrt{x}}{\sqrt{x}-1}\cdot\dfrac{1}{\sqrt{x}}=\dfrac{2}{\sqrt{x}-1}\)

Tick hộ nha

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

\(A=1-\dfrac{2\left(2\sqrt{x}-1\right)-5\sqrt{x}+\left(2\sqrt{x}+1\right)}{\left(2\sqrt{x}-1\right)\left(2\sqrt{x}+1\right)}:\dfrac{\sqrt{x}-1}{\left(2\sqrt{x}+1\right)^2}\)

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

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

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

Tick hộ nha

a. P: Cây hoa đỏ ---> F1: cây hoa trắng => tính trạng hoa trắng là tính trạng lặn.  
Quy ước: gen A - hoa đỏ, gen a - hoa trắng  
=> cây hoa trắng F1 là aa => cây hoa đỏ P: Aa  
Ta có:  
P: Aa x Aa  
Gp: 1A; 1a  
F1: 1AA: 2Aa: 1aa (3 cây hoa đỏ: 1 cây hoa trắng).

???

Xét tam giác ABC có \(\widehat{BAC}+\widehat{ACB}+\widehat{B}=180\Rightarrow\widehat{BAD}+\widehat{DAC}+\widehat{ACD}+\widehat{BAD}+\widehat{B}=180\) độ

hay \(20+\widehat{DAC}+25+\widehat{ACD}+50=180\) độ

\(\Rightarrow\widehat{DAC}+\widehat{ACD}=85\) độ

Xét tam giác ADC có \(\widehat{DAC}+\widehat{ACD}+\widehat{ADC}=180\Rightarrow\widehat{ADC}=180-85=95\) độ

Tick hộ nha

Đặt \(\sqrt{1+x}=a;\sqrt{1-x}=b\), \(a,b>0\)

Áp dụng BĐT AG-GM:

\(\Rightarrow A=\dfrac{a^2+4b^2}{ab}\ge\dfrac{2\sqrt{a^2\cdot4b^2}}{ab}=4\)

Dấu "=" \(\Leftrightarrow1+x=4\left(1-x\right)\Leftrightarrow x=\dfrac{3}{5}\left(N\right)\)

Tick hộ nha

Đặt \(A=\sqrt[3]{22\sqrt{2}+25}-\sqrt[3]{22\sqrt{2}-25}\)

\(\Rightarrow A^3=50-3\sqrt[3]{\left(22\sqrt{2}+25\right)\left(22\sqrt{2}-25\right)}\left(\sqrt[3]{22\sqrt{2}+25}-\sqrt[3]{22\sqrt{2}-25}\right)\)

\(\Rightarrow A^3=50-3\sqrt[3]{\left(22\sqrt{2}+25\right)\left(22\sqrt{2}-25\right)}\cdot A\)

\(\Rightarrow A^3=50-3A\sqrt[3]{343}=50-21A\)

\(\Rightarrow A^3+21A-50=0\Leftrightarrow A^3-4A+25A-50=0\)

\(\Leftrightarrow\left(A-2\right)\left(A^2+2A+25\right)=0\)

\(\Leftrightarrow A=2\left(A^2+2A+25>0,\forall A\right)\)

\(\Rightarrow\sqrt[3]{22\sqrt{2}+25}-\sqrt[3]{22\sqrt{2}-25}=2\)

Tick nha bạn 😘

Xét \(VT=a+2b+c=1+b\left(1\right)\)

Áp dụng BĐT AG-GM:

\(4\left(1-a\right)\left(1-c\right)\le\left(1-a+1-c\right)^2=\left(2-a-c\right)^2=\left(1+a+b+c-a-c\right)^2=\left(1+b\right)^2\left(2\right)\)

\(\Rightarrow4\left(1-a\right)\left(1-b\right)\left(1-c\right)\le\left(1-b\right)\left(1+b\right)^2\)

Mà \(\left(1-b\right)\left(1+b\right)^2-\left(1-b\right)=\left(1+b\right)\left(1-b^2-1\right)=-b^2\left(1+b\right)\le0,\forall b\ge0\)

Do đó \(\left(1-b\right)\left(1+b\right)^2\le1+b\left(3\right)\)

Từ \(\left(1\right)\left(2\right)\left(3\right)\) ta có ĐPCM

Dấu "=" \(\Leftrightarrow a=c=\dfrac{1}{2};b=0\) 

Ta có \(\left(\sqrt{2018}+\sqrt{2020}\right)^2=4038+2\sqrt{4076360}\) và \(\left(2\sqrt{2019}\right)^2=8076=4038+4038\)

Mà \(\left(2\sqrt{4076360}\right)^2=16305440\) và \(4038^2=16305444\)

\(\Rightarrow2\sqrt{4076360}< 4038\)

\(\Rightarrow\sqrt{2018}+\sqrt{2020}< 2\sqrt{2019}\)

tui đổi mà cũng ko đc

Để \(x^2+mx-8⋮x+1\)

\(\Leftrightarrow x^2+mx-8=\left(x+1\right).A\left(x\right)\)

Thay \(x=-1\)

\(\Leftrightarrow1-m-8=0\Leftrightarrow m=7\)