One day. as a farmer was in his field and his buffalo (1)_WAS GRAZING__ nearby, a tiger (2)_APPEARED__. The tiger wanted to know why the strong buffalo was the servant and the small man (3)__WAS_ the master. The farmer (4)_SAID__ he had something called wisdom, but he (5)_LEFT__ it at home that day. He (6)_WENT__ to get the wisdom, but before that he (7)__TIED_ the tiger to a tree with a rope because he didn't want it to eat the buffalo. When he returned, the farmer brought some straw with him. He said it was his wisdom. He (8)__LIT_ the straw and the fire (8)_BURNED__ the tiger. The tiger (9)_ESCAPED__, but it still has black stripes from the burns today.

a) \(F=2\left|3x-2\right|-1\)

Vì \(\left|3x-2\right|\ge0\forall x\Rightarrow2\left|3x-2\right|\ge0\)

\(\Rightarrow2\left|3x-2\right|-1\ge-1\)

''='' xảy ra khi \(3x-2=0\Rightarrow x=\dfrac{2}{3}\)

=> \(F_{min}=-1\)

b) \(G=x^2+3\left|y-2\right|-1\)

Ta có: \(\left\{{}\begin{matrix}x^2\ge0\forall x\\3\left|y-2\right|\ge0\forall y\end{matrix}\right.\)

=> \(x^2+3\left|y-2\right|\ge0\Rightarrow x^2+3\left|y-2\right|-1\ge-1\)

''='' xảy ra khi \(\left\{{}\begin{matrix}x^2=0\\y-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\y=2\end{matrix}\right.\)

Vậy \(G_{min}=-1\)

Bài 3:

d)D = \(\sqrt{-x^2+2x-1}=\sqrt{-\left(x-1\right)^2}\)

\(-\left(x-1\right)^2\le0\forall x\)

để D có nghĩa => \(x-1=0\Rightarrow x=1\)

c) \(C=\sqrt{9x^2-6x+1}=\sqrt{\left(3x-1\right)^2}\)

Vì \(\left(3x-1\right)^2\ge0\forall x\) => căn thức có nghĩa với mọi x \(\in R\)

P/s: Đánh dấu 2 ý này --> lm 2 ý

Sửa đề x-y+z = -85 (số này đẹp hơn :))

Áp dụng t/c của dãy tỉ số = nhau có:

\(\dfrac{x+3}{5}=\dfrac{y-2}{3}=\dfrac{z-1}{7}=\dfrac{x+3-y+2+z-1}{5-3+7}=\dfrac{-85+4}{9}=-9\)

\(\Rightarrow\left\{{}\begin{matrix}x+3=-45\\y-2=-27\\z-1=-63\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-48\\y=-25\\z=-62\end{matrix}\right.\)

Vậy...............

P/s: Nếu đề bạn đúng thì thay số vào thôi nhé!

Giải:

a) Xét \(\Delta ABO\) và \(\Delta ECO\) có:

\(\widehat{ABO}=\widehat{OCE}\) (=90^o)

\(BO=CO\left(gt\right)\)

\(\widehat{O_1}=\widehat{O_2}\)

=> \(\Delta ABO=\Delta ECO\) (g.c.g)

=> \(AO=EO\)

Tứ giác ABEC có: \(\left\{{}\begin{matrix}AO=EO\left(cmt\right)\\BO=CO\left(gt\right)\end{matrix}\right.\) => O là trung điểm của AE và BC

=> ABEC là hbh (đpcm)

b) Vì AB = CD (ABCD là hcn) => CD = CE => C là trung điểm của DE (1)

mặt \(\ne\) C là trung điểm của BF (2)

=> BEFD là hbh (*)

lại có BE = BD (cùng = AC) (**)

Từ (*) và (**) => BEFD là hình thoi (đpcm)

c) Có: \(\widehat{C_2}+\widehat{C_3}+\widehat{C_4}=180^o\)

mà \(\widehat{C_1}=\widehat{C_2}\) (đối đỉnh)

=> \(\widehat{C_1}+\widehat{C_3}+\widehat{C_4}=180^o\)

hay \(\widehat{ACI}=180^o\) => 3 điểm A, C, I thẳng hàng

\(x^8+x+1=x^8+x^7-x^7+x^6-x^6+x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x+1\)

\(=\left(x^8+x^7+x^6\right)-\left(x^7+x^6+x^5\right)+\left(x^5+x^4+x^3\right)-\left(x^4+x^3+x^2\right)+\left(x^2+x+1\right)\)

\(=x^6\left(x^2+x+1\right)-x^5\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)

Ta có: \(7^8\equiv426\left(mod2005\right)\)

\(\left(7^8\right)^2\equiv426^2\equiv1026\left(mod2005\right)\)

\(\Rightarrow7^{17}\equiv7^{16}\cdot7\equiv1026\cdot7\equiv7182\equiv1167\left(mod2005\right)\)