,m mmkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkkbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbbb

11. They won’t come here again.

-> I wish that they wouldn't come here again

12. He won’t go swimming with me.

-> I wish that he wouldn't go swimming with me 

13. I will be late for school.

->  I wish that I wouldn't be late for school

jjjjjjjjjjjjjjjjjjj

lực cười

cười một mình đi 🤪

a² + b² + c² = ab + bc + ca

<=> 2a² + 2b² + 2c² = 2ab + 2bc + 2ca

<=> 2a² + 2b² + 2c² - 2ab - 2bc - 2ca = 0

<=> (a² - 2ab + b²) + (a² - 2ca + c²) + (b² - 2bc + c²) = 0

<=> (a - b)² + (a - c)² + (b - c)² = 0

<=> \(\hept{\begin{cases}a-b=0\\a-c=0\\b-c=0\end{cases}}\Leftrightarrow a=b=c\)(đpcm) 

Nhân vế 2 biểu thức, ta có:

\(2a^2+2b^2+2c^2=2ab+2bc+2ca\)

\(\Leftrightarrow2a^2+2b^2+2c^2-2ab+2bc+2ca=0\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(b^2-2bc+c^2\right)+\left(c^2-2ac+a^2\right)=0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\left(1\right)\)

Vì\(\left(a-b\right)^2\ge0;\left(b-c\right)^2\ge0;\left(c-a\right)^2\ge0\)nên từ (1) \(\Rightarrow a-b=b-c=c-a=0\)hay \(a=b=c\)

Ta có:\(\frac{1}{ab}+\frac{1}{cd}\ge\frac{8}{\left(a+b\right)\left(c+d\right)}\Leftrightarrow\left(\frac{1}{ab}+\frac{1}{cd}\right)\left(a+b\right)\left(c+d\right)\ge8\)

Xét bất đẳng thức Cô si

\(\hept{\begin{cases}\frac{1}{ab}+\frac{1}{cd}\ge2\sqrt{\frac{1}{abcd}}\\a+b\ge2\sqrt{ab}\\c+d\ge2\sqrt{cd}\end{cases}}\)

\(\Rightarrow\left(\frac{1}{ab}+\frac{1}{cd}\right)\left(a+b\right)\left(c+d\right)\ge2\cdot\frac{1}{\sqrt{abcd}}\cdot2\sqrt{ab}\cdot2\sqrt{cd}\)

\(\Rightarrow\left(\frac{1}{ab}+\frac{1}{cd}\right)\left(a+b\right)\left(c+d\right)\ge8\left(đpcm\right)\)

Dấu "=" xảy ra khi\(\hept{\begin{cases}\frac{1}{ab}=\frac{1}{cd}\\a=b\\c=d\end{cases}}\Leftrightarrow a=b=c=d\)

Mình thì dư đoán điểm rơi \(a=b=c=1\) rồi, nhưng nháp mãi vẫn không ra được.

\(\frac{a}{b^3+ab}\)=\(\frac{a^2}{b^3a+a^2b}\)

tương tự thì ta có S= \(\frac{a^2}{b^3a+a^2b}\) +     \(\frac{b^2}{c^3b+b^2c}\)   +    \(\frac{c^2}{a^3c+ac^2}\)

áp dụng bất dẳng thức cô si s goát,ta có

S=\(\frac{a^2}{b^3a+a^2b}\)+     \(\frac{b^2}{c^3b+b^2c}\)+    \(\frac{c^2}{a^3c+ac^2}\)\(\ge\)   \(\frac{\left(a+b+c\right)^2}{b^3a+a^2b+c^3b+b^2c+a^3c+c^2a}\)

cái mẫu mk chx nghĩ  ra phân tích ra sao nx,tí nghĩ nốt

bí rùi

6 nha 

HT

b) Ta có \(A=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\ge\frac{\left(x+y+z\right)^2}{y+z+z+x+x+y}\)(BĐT Schwarz) 

\(=\frac{x+y+z}{2}=\frac{2}{2}=1\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}\frac{x^2}{y+z}=\frac{y^2}{z+x}=\frac{z^2}{x+y}\\x+y+z=2\end{cases}}\Leftrightarrow x=y=z=\frac{2}{3}\)

a) Có \(P=1.\sqrt{2x+yz}+1.\sqrt{2y+xz}+1.\sqrt{2z+xy}\)

\(\le\sqrt{\left(1^2+1^2+1^2\right)\left(2x+yz+2y+xz+2z+xy\right)}\)(BĐT Bunyakovsky) 

\(=\sqrt{3.\left[2\left(x+y+z\right)+xy+yz+zx\right]}\)

\(\le\sqrt{3\left[4+\frac{\left(x+y+z\right)^2}{3}\right]}=\sqrt{3\left(4+\frac{4}{3}\right)}=4\)

Dấu "=" xảy ra <=> x = y = z = 2/3 

Part 1.

1 B

2 A

3 A

4 C

5 B