sao ko thay anh đi

Read the following passage and blacken the letter A, B, C, or D on your answer sheet to indicate the correct answer to each of the questions.

        Sharks have gained an unfair reputation for being fierce predators of large sea animals. Humanity's unfounded fear and hatred of these ancient creatures is leading to a worldwide slaughter that may result in the extinction of many coastal shark species. The shark is the victim of a warped attitude of wildlife protection; we strive only to protect the beautiful, non-threatening parts of our environment. And, in our efforts to restore only non-threatening parts of our earth, we ignore other important parts.

        A perfect illustration of this attitude is the contrasting attitude toward another large sea animal, the dolphin. During the 1980s, environmentalists in the United States protested the use of driftnets for tuna fishing in the Pacific Ocean since these nets also caught dolphins. The environmentalists generated enough political and economic pressure to prevent tuna companies from buying tuna that had been caught in driftnets. In contrast to this effort on behalf of the dolphins, these same environmentalists have done very little to help save the Pacific Ocean sharks whose population has decreased nearly to the point of extinction.

Sharks are among the oldest creatures on earth, having survived in the seas for more than 350 million years. They are extremely efficient animals, feeding on wounded or dying animals, thus performing an important role in nature of weeding out the weaker animals in a species. Just the fact that species such as the Great White Shark have managed to live in the oceans for so many millions of years is enough proof of their efficiency and adaptability to changing environments. It is time for US humans, who may not survive another 1,000 years at the rate we are damaging the planet, to cast away our fears and begin considering the protection of sharks as an important part of a program for protection of all our natural environment.

How did environmentalists manage to protect dolphins?

A. They prevented fishermen from selling them for meat.

B. They pressured fishermen into protecting dolphins by law.

C. They brought political pressure against tuna companies.

D. They created sanctuaries where dolphin fishing was not allowed.

1/\(A=\dfrac{x^2-2x+2014}{x^2}\)

\(\Leftrightarrow A=\dfrac{2014x^2-2.x.2014+2014^2}{2014x^2}\)

\(\Leftrightarrow A=\dfrac{2013x^2+x^2-2.x.2014+2014^2}{2014x^2}\)

\(\Leftrightarrow A=\dfrac{2013x^2+\left(x-2014\right)^2}{2014x^2}\)

\(\Leftrightarrow A=\dfrac{2013}{2014}+\dfrac{\left(x-2014\right)^2}{2014x^2}\)

Có: \(\left(x-2014\right)^2\ge0\forall x\)

\(2014x^2>0\forall xvìx\ne0\)

\(\Rightarrow\dfrac{\left(x-2014\right)^2}{2014x^2}\ge0\)

\(\Rightarrow\dfrac{2013}{2014}+\dfrac{\left(x-2014\right)^2}{2014x^2}\ge\dfrac{2013}{2014}\)

\(\Rightarrow A\ge\dfrac{2013}{2014}\)

dấu "=" xảy ra khi và chỉ khi x - 2014 =0 <=> x = 2014

Vậy \(min_A=\dfrac{2013}{2014}\Leftrightarrow x=2014\)

2) Ta có:

\(x=\sqrt{a+\sqrt{a^2-1}}+\sqrt{a-\sqrt{a^2-1}}\)

\(\Leftrightarrow x^2=a-\sqrt{a^2-1}+2\sqrt{a-\sqrt{a^2-1}}.\sqrt{a+\sqrt{a^2-1}}+a+\sqrt{a^2-1}\)

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

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

\(\Leftrightarrow x^2=2a+2=2\left(a+1\right)\)

\(\Leftrightarrow-x^3=-2\left(a+1\right)x\)

Đặt \(A=x^3-2x^2-2\left(a+1\right)x+4x+2021\)

\(\Leftrightarrow A=x^3-2\left(2a+2\right)-x^3+4a+2021\)

\(\Leftrightarrow A=-4a-4+4a+2021\)

\(\Leftrightarrow A=2017\)

\(\sqrt{x+5}-y^3=\sqrt{y+5}-x^3\)

\(\Leftrightarrow\sqrt{x+5}-\sqrt{y+5}+x^3-y^3=0\)

\(\Leftrightarrow\dfrac{\left(\sqrt{x+5}-\sqrt{y+5}\right)\left(\sqrt{x+5}+\sqrt{y+5}\right)}{\sqrt{x+5}+\sqrt{y+5}}+\left(x-y\right)\left(x^2+xy+y^2\right)=0\)

\(\Leftrightarrow\dfrac{\left(x+5\right)-\left(y+5\right)}{\sqrt{x+5}+\sqrt{y+5}}+\left(x-y\right)\left(x^2+xy+y^2\right)=0\)

\(\Leftrightarrow\dfrac{x-y}{\sqrt{x+5}+\sqrt{y+5}}+\left(x-y\right)\left(x^2+xy+y^2\right)=0\)

\(\Leftrightarrow\left(x-y\right)\left(\dfrac{1}{\sqrt{x+5}+\sqrt{y+5}}+x^2+xy+y^2\right)=0\)

\(\Leftrightarrow x-y=0\) và\(\dfrac{1}{\sqrt{x+5}+\sqrt{y+5}}+x^2+xy+y^2>0\)

\(\Leftrightarrow x=y\)

Khi đó: \(P=x^2-3xy+12-y^2+2018\)

\(\Leftrightarrow P=x^2-3x^2+12-x^2+2018\)

\(\Leftrightarrow P=2030-3x^2\le2030\)

dấu "=" xảy ra khi và chỉ khi x=y=0

vậy maxP=2030 khi và chỉ khi x=y=0

Bạn tự vẽ hình nha:

a) Có AD, BE, CF là các đường cao của tam giác ABC

=>\(AD\perp BC;BE\perp AC;CF\perp AB\)

\(\widehat{BFC}=\widehat{BEC}=90^o\)

tứ giác BFEC có F và E là 2 đỉnh liên tiếp cùng nhìn đoạn BC dưới góc 90^okhông đổi.

=> tứ giác BFEC nội tiếp.

tứ giác DHEC có \(\widehat{HDC}+\widehat{HEC}=90^o+90^o=180^o\)

=> tứ giác DHEC nội tiếp.

b) tứ giác BFEC nội tiếp (cmt)

=> \(\widehat{FEH}=\widehat{NBC}\)(cùng chắn cung BF)

lại có : \(\widehat{NBC}=\widehat{NMH}\)( 2 góc nội tiếp cùng chắn cung BN)

=> góc FEH = góc NMH

mà 2 góc này ở vị trí đồng vị

=>\(\)EF//MN.

c) kẻ đường kính AK

=>\(\widehat{ACK}=90^o\)( góc nội tiếp chắn nửa đường tròn)

\(\Delta ABD\) và \(\Delta AKC\) có:

\(\widehat{ABD}=\widehat{AKC}\)(cùng chắn cung AC)

\(\widehat{ADB}=\widehat{ACK}=90^o\)

=> \(\Delta ABD\sim\Delta AKC\left(g.g\right)\)

\(\Rightarrow\dfrac{AB}{AK}=\dfrac{AD}{AC}\)

=> AB.AC=AK.AD

=>AB.AC=2R.AD

Hoành độ giao điểm của (d) và (P) là nghiệm của phương trình:

\(x^2=ax+3\)

\(\Leftrightarrow x^2-ax-3=0\)

có \(\Delta=\left(-a\right)^2-4.1.\left(-3\right)=a^2+12\ge12>0\forall a\)

=> phương trình luôn có 2 nghiệm phân biệt x1,x2

theo hệ thức Vi-ét ta có:

\(\left\{{}\begin{matrix}x_1+x_2=a\left(1\right)\\x_1x_2=-3\left(2\right)\end{matrix}\right.\)

mà \(x_1+2x_2=3\left(3\right)\)

Từ (1) và (3) ta có hệ:

\(\left\{{}\begin{matrix}x_1+x_2=a\\x_1+2x_2=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_2=3-a\\x_1+x_2=a\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x_1=2a-3\\x_2=3-a\end{matrix}\right.\)

Theo (2) ta có:

\(x_1x_2=-3\Leftrightarrow\left(2a-3\right)\left(3-a\right)=-3\)

\(\Leftrightarrow6a-2a^2-9+3a=-3\)

\(\Leftrightarrow2a^2-9a+6=0\)

\(\Delta=\left(-9\right)^2-4.2.6=81-48=33>0\)

\(\Rightarrow a_1=\dfrac{9+\sqrt{33}}{4};a_2=\dfrac{9-\sqrt{33}}{4}\)

Vậy \(a\in\left\{\dfrac{9+\sqrt{33}}{4};\dfrac{9-\sqrt{33}}{4}\right\}\)để x1+2x2=3

8/33 tik mk nha

tìm bội chung của 6; 8; 7;15 === 

300 : 100 x 3= 9

duyệt nha