\(a=\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{1+\dfrac{1}{x^2}}+1}{\dfrac{5}{x}-2}=\dfrac{1+1}{-2}=-1\)

Câu b bạn coi lại đề, số cuối là \(-3\) hay \(-7\) nhỉ? 

Kí hiệu lim làm theo thứ tự này:

Câu b số cuối là -3 thì nó hơi kì quặc (vì không phải dạng vô định, cứ thay số thôi):

\(=\lim\limits_{x\rightarrow4}\dfrac{\sqrt{15.4+4}-\sqrt{4-3}-3}{-4+4}=\dfrac{4}{0}=+\infty\)

\(ĐK:x\ge0\\ PT\Leftrightarrow\dfrac{5}{3}\sqrt{15x}-\sqrt{15x}-\dfrac{1}{3}\sqrt{15x}=2\\ \Leftrightarrow\sqrt{15x}\left(\dfrac{5}{3}-1-\dfrac{1}{3}\right)=2\\ \Leftrightarrow\dfrac{1}{3}\sqrt{15x}=2\Leftrightarrow\sqrt{15x}=6\Leftrightarrow15x=36\\ \Leftrightarrow x=\dfrac{12}{5}\left(tm\right)\)

\(ĐK:x\ge0\\ PT\Leftrightarrow\dfrac{2}{3}\sqrt{15x}-\dfrac{1}{3}\sqrt{15x}=2\\ \Leftrightarrow\dfrac{1}{3}\sqrt{15x}=2\Leftrightarrow\sqrt{15x}=6\\ \Leftrightarrow15x=36\Leftrightarrow x=\dfrac{12}{5}\left(tm\right)\)

\(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\)

\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)

\(\Leftrightarrow4\sqrt{x+1}=16\)

\(\Leftrightarrow\sqrt{x+1}=4\)

<=> x + 1 = 16

<=> x = 15 (nhận)

~ ~ ~

\(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=6\)

\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)

\(\Leftrightarrow3\sqrt{x+5}=6\)

\(\Leftrightarrow\sqrt{x+5}=2\)

<=> x + 5 = 4

<=> x = - 1 (nhận)

tính tan40°×tan45°×tan50°
#Help me -.-

pt j ạ

Đặt \(\sqrt{15x}=a\)

Pt sẽ là \(\dfrac{5}{3a}-a+11=\dfrac{1}{3a}\)

=>\(\dfrac{4}{3a}=a-11\)

\(\Leftrightarrow3a^2-33a-4=0\)

=>\(a=11.12\)

=>căn 15x=11,12

=>15x=123,6544

hay \(x\simeq8,24\)

3.

Ta có: \(VT=\)\(8+2\sqrt{10+2\sqrt{5}}+8-2\sqrt{10+2\sqrt{5}}\)

\(=8+8+\left(2\sqrt{10+2\sqrt{5}}-2\sqrt{10+2\sqrt{5}}\right)\)

\(=16\ne VP\)

⇒ Đề sai

1. Ta có: \(\sqrt{4x}\)- 3\(\sqrt{x}\)+2\(\sqrt{15x}\)=18

⇌2\(\sqrt{x}\)-3\(\sqrt{x}\) +2\(\sqrt{15x}\)=18

⇌\(-\sqrt{x}\) +2\(\sqrt{15x}\)-15 = 3

⇌-(\(\sqrt{x}\) -2\(\sqrt{15x}\)+15 )=3

⇌(\(\sqrt{x}\)-\(\sqrt{15}\))=-3 (vô lí)

Vậy không tìm được giá trị x thỏa mãn bài toán

2.Ta có: B=\(\dfrac{1}{\sqrt{11-2\sqrt{30}}}-\dfrac{3}{7-2\sqrt{10}}\)

= \(\dfrac{1}{\sqrt{6-2\sqrt{6.5}+5}}-\dfrac{3}{2-2\sqrt{2.5}+5}\)

=\(\dfrac{1}{\sqrt{\left(\sqrt{6}-\sqrt{5}\right)^2}}-\dfrac{3}{\left(\sqrt{3}-\sqrt{2}\right)^2}\)

=\(\dfrac{1}{\sqrt{6}-\sqrt{5}}-\dfrac{3}{\sqrt{3}-\sqrt{2}}\)

hình như đề sai

Câu 1: 

1: Ta có: \(P=\left(\dfrac{x^2}{x^2-3}+\dfrac{2x^2-24}{x^4-9}\right)\cdot\dfrac{7}{x^2+8}\)

\(=\left(\dfrac{x^2\left(x^2+3\right)}{\left(x^2-3\right)\left(x^2+3\right)}+\dfrac{2x^2-24}{\left(x^2-3\right)\left(x^2+3\right)}\right)\cdot\dfrac{7}{x^2+8}\)

\(=\dfrac{x^4+3x^2+2x^2-24}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)

\(=\dfrac{x^4+5x^2-24}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)

\(=\dfrac{x^4+8x^2-3x^2-24}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)

\(=\dfrac{x^2\left(x^2+8\right)-3\left(x^2+8\right)}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)

\(=\dfrac{\left(x^2+8\right)\left(x^2-3\right)}{\left(x^2-3\right)\left(x^2+3\right)}\cdot\dfrac{7}{x^2+8}\)

\(=\dfrac{7}{x^2+3}\)

Câu 2a đề sai, pt này ko giải được

2b.

\(P\left(x\right)=\left(2x+7\right)\left(x^2-4x+4\right)+\left(a+20\right)x+\left(b-28\right)\)

Do \(\left(2x+7\right)\left(x^2-4x+4\right)⋮\left(x^2-4x+4\right)\)

\(\Rightarrow P\left(x\right)\) chia hết \(Q\left(x\right)\) khi \(\left(a+20\right)x+\left(b-28\right)\) chia hết \(x^2-4x+4\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+20=0\\b-28=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=-20\\b=28\end{matrix}\right.\)

3a.

\(VT=\dfrac{1}{1+x^2}+\dfrac{1}{1+y^2}=\dfrac{2+x^2+y^2}{1+x^2+y^2+x^2y^2}=1+\dfrac{1-x^2y^2}{1+x^2+y^2+x^2y^2}\le1+\dfrac{1-x^2y^2}{1+2xy+x^2y^2}\)

\(VT\le1+\dfrac{\left(1-xy\right)\left(1+xy\right)}{\left(xy+1\right)^2}=1+\dfrac{1-xy}{1+xy}=\dfrac{2}{1+xy}\) (đpcm)

3b

Ta có: \(n^3-n=n\left(n-1\right)\left(n+1\right)\) là tích 3 số nguyên liên tiếp nên luôn chia hết cho 6

\(\Rightarrow n^3\) luôn đồng dư với n khi chia 6

\(\Rightarrow S\equiv2021^{2022}\left(mod6\right)\)

Mà \(2021\equiv1\left(mod6\right)\Rightarrow2021^{2020}\equiv1\left(mod6\right)\)

\(\Rightarrow2021^{2022}-1⋮6\)

\(\Rightarrow S-1⋮6\)

2a. 

À nãy mình nhìn lộn dấu trừ bên vế phải thành dấu cộng

ĐKXĐ: ...

\(\Leftrightarrow\dfrac{3x+2022+2x-2021}{\left(2x-2021\right)\left(3x+2022\right)}=\dfrac{10x-2024-\left(15x-2023\right)}{\left(15x-2023\right)\left(10x-2024\right)}\)

\(\Leftrightarrow\dfrac{5x-1}{\left(2x-2021\right)\left(3x+2022\right)}=-\dfrac{5x-1}{\left(15x-2023\right)\left(10x-2024\right)}\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-1=0\Rightarrow x=...\\\dfrac{1}{\left(2x-2021\right)\left(3x+2022\right)}=-\dfrac{1}{\left(15x-2023\right)\left(10x-2024\right)}\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\left(2x-2021\right)\left(3x+2022\right)+\left(15x-2023\right)\left(10x-2024\right)=0\)

\(\Leftrightarrow\left[12x-4045-\left(10x-2024\right)\right]\left(3x+2022\right)+\left(12x-4045+3x+2022\right)\left(10x-2024\right)=0\)

\(\Leftrightarrow\left(12x-4045\right)\left(3x+2022\right)-\left(10x-2024\right)\left(3x+2022\right)+\left(12x-4045\right)\left(10x-2024\right)+\left(3x+2022\right)\left(10x-2024\right)=0\)

\(\Leftrightarrow\left(12x-4045\right)\left(13x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{13}\\x=\dfrac{4045}{12}\end{matrix}\right.\)

chắc bạn chép sai đề rồi , hai căn đầu phải 1 cộng 1 trừ chứ

Đặt

\(x=\dfrac{y+1}{2}\Rightarrow y=2x-1\)

\(\Rightarrow y=\sqrt[3]{4+\sqrt{15}}+\sqrt[3]{4-\sqrt{15}}\)

\(y^3=8+3\sqrt[3]{\left(4-\sqrt{15}\right)\left(4+\sqrt{15}\right)}=8+3y\)

\(\Rightarrow y^3-3y-8=0\\ \)

\(\Leftrightarrow8x^3-12x^2-6=0\)

\(\Rightarrow4x^3-6x^2-3=0\)

thay p vào ta có

\(P=12x^5-18x^4+4x^3-15x^2-21\)

\(=12x^5-18x^4-9x^2-4x^3-6x^2-21\)

\(=3x^2\left(4x^2-6x^2-3\right)+4x^3-6x^2-3\\ =3x^2.0+0-18\\ =-18\)