Ta có: \(M=\left(2x-1\right)\left(2y-1\right)\)

\(=4xy-2x-2y+1\)

\(=4xy-2\left(x+y\right)+1\)

Thay x + y = 10 và xy = 16

\(\Rightarrow M=64-20+1=45\)

Vậy M = 45

Ta có: \(\left(-32\right)^9=-32^9\)

\(=-\left(2^5\right)^9=-2^{45}\)

\(\left(-16\right)^{13}=-\left(2^4\right)^{13}=-2^{52}\)

Vì \(-2^{45}>-2^{52}\Rightarrow\left(-32\right)^9>\left(-16\right)^{13}\)

Vậy...

Đề bài là gì và bạn kiểm tra lại xem sai đề không nhé!

\(\left(1-\sqrt{x}\right)\left(1+\sqrt{x}+x\right)=1-\sqrt{x}^3\)

Áp dụng bất đẳng thức \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) có:

\(A=\left|x-3\right|+\left|x+\dfrac{3}{2}\right|=\left|3-x\right|+\left|x+\dfrac{3}{2}\right|\)

\(\ge\left|3-x+x+\dfrac{3}{2}\right|=\left|\dfrac{9}{2}\right|=\dfrac{9}{2}\)

Dấu " = " khi \(\left\{{}\begin{matrix}3-x\ge0\\x+\dfrac{3}{2}\ge0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\le3\\x\ge\dfrac{-3}{2}\end{matrix}\right.\)

Vậy \(MIN_A=\dfrac{9}{2}\) khi \(\dfrac{-3}{2}\le x\le3\)

1. An elephant is _bigger than__ a mouse.

2. The weather today is _hotter than__ it was yesterday.

3. A diamond costs a lot of money. A diamond is _more expensive than__ a ruby.

4. A lake is _smaller than__ an ocean.

5. A person can think logically. A person is _more intelligent than__ an animal.

6. Good health is _more important than__ money.

7. I can buy a bicycle, but not a motorbike. A bicycle is _cheaper than__ a motorbike.

8. The last question is _harder than__ the others.

9. I think my second essay is _worse than__ the first. There were many mistakes in the first essay.

10. The food in a street market is _better than__ that in a supermarket.

Giải:

a, Ta có: \(\widehat{ADE}=\widehat{BAD}\) ( so le trong do ED // AB )

Mà \(\widehat{BAD}=\widehat{CAD}\left(gt\right)\)

\(\Rightarrow\widehat{ADE}=\widehat{CAD}\)

\(\Rightarrowđpcm\)

b, Nối F với D

Xét \(\Delta BFD,\Delta EDF\) có:

\(\widehat{BFD}=\widehat{EDF}\) ( so le trong do EF // BC )

FD: cạnh chung

\(\widehat{BDF}=\widehat{EFD}\) ( so le trong do EF // BC )

\(\Rightarrow\Delta BFD=\Delta EDF\left(g-c-g\right)\)

\(\Rightarrow\widehat{FBD}=\widehat{DEF}\) ( góc t/ứng )

hay \(\widehat{ABC}=\widehat{DEF}\)

\(\Rightarrowđpcm\)