Đáp án: C

Giải thích: Mục…4….Trang…22…..SGK Lịch sử 11 cơ bản

Câu trả lời bằng hình ảnh

Program OLM;

var  a,b: integer;

s,p: longint;

begin

write('Nhap chieu dai: '); readln(a);

write('Nhap chieu rong: '); readln(b);

s:=a*b;

p:=(a+b)*2;

writeln('Chu vi hinh chu nhat la: ',p); 

writeln('Dien tich hinh chu nhat la: ',s); 

readln

end.

Biết SA và DA làm sao nhỉ? Đề bị lỗi

\(\lim\limits_{x\rightarrow1}\dfrac{\sqrt[]{4x-3}-\sqrt[3]{6x-5}}{x^3-x^2-x+1}=\lim\limits_{x\rightarrow1}\dfrac{\left(\sqrt[]{4x-3}-\left(2x-1\right)\right)+\left(\left(2x-1\right)-\sqrt[3]{6x-5}\right)}{x^2\left(x-1\right)-\left(x-1\right)}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{4x-3-\left(2x-1\right)^2}{\sqrt[]{4x-3}+2x-1}+\dfrac{\left(2x-1\right)^3-\left(6x-5\right)}{\left(2x-1\right)^2+\left(2x-1\right)\sqrt[3]{6x-5}+\sqrt[3]{6x-5}}}{\left(x-1\right)^2\left(x+1\right)}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{\dfrac{-4\left(x-1\right)^2}{\sqrt[]{4x-3}+2x-1}+\dfrac{4\left(x-1\right)^2\left(2x+1\right)}{\left(2x-1\right)^2+\left(2x-1\right)\sqrt[3]{6x-5}+\sqrt[3]{\left(6x-5\right)^2}}}{\left(x-1\right)^2\left(x+1\right)}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{-\dfrac{4}{\sqrt[]{4x-3}+2x-1}+\dfrac{4\left(2x+1\right)}{\left(2x-1\right)^2+\left(2x-1\right)\sqrt[3]{6x-5}+\sqrt[3]{\left(6x-5\right)^2}}}{x+1}=1\)