Cách làm ngắn gọn: \(5=\dfrac{5\left(x-1\right)}{x-1}=\dfrac{5x-5}{x-1}=\dfrac{5x+5-10}{x-1}\)

Do đó chọn \(f\left(x\right)=5x+5\) thế vào nhanh chóng tính ra kết quả giới hạn

Còn cách khác phức tạp hơn (có thể sử dụng cho tự luận):

Do \(\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-10}{x-1}=5\) hữu hạn nên \(f\left(x\right)-10=0\) có nghiệm \(x=1\)

\(\Rightarrow f\left(1\right)-10=0\Rightarrow f\left(1\right)=10\)

Do đó:

\(\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-10}{\left(\sqrt{x}-1\right)\left(\sqrt{4f\left(x\right)+9}+3\right)}=\lim\limits_{x\rightarrow1}\dfrac{\left[f\left(x\right)-10\right]\left(\sqrt{x}+1\right)}{\left(x-1\right)\left(\sqrt{4f\left(x\right)+9}+3\right)}\)

\(=\lim\limits_{x\rightarrow1}\dfrac{f\left(x\right)-10}{x-1}.\dfrac{\sqrt{x}+1}{\sqrt{4f\left(x\right)+9}+3}=5.\dfrac{1+1}{\sqrt{4f\left(1\right)+9}+3}=5.\dfrac{2}{\sqrt{4.10+9}+3}=...\)

\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{40.43}+\frac{1}{43.46}\)

\(=3.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{43}-\frac{1}{46}\right)\)

\(=3.\left(1-\frac{1}{46}\right)\)

\(=3.\frac{45}{46}\)

\(=\frac{135}{46}\)

~Học tốt~

\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{40.43}+\frac{3}{43.46}\)

\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{40}-\frac{1}{43}+\frac{1}{43}-\frac{1}{46}\)

\(=1-\frac{1}{46}\)

\(=\frac{45}{46}\)

~ Hok tốt ~

Bài 6:

a: ĐKXĐ: x>=0; x<>4

\(D=\left(\frac{\sqrt{x}}{x-4}+\frac{2}{2-\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\frac{10-x}{\sqrt{x}+2}\right)\)

\(=\frac{\sqrt{x}-2\left(\sqrt{x}+2\right)+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\frac{x-4+10-x}{\sqrt{x}+2}\)

\(=\frac{2\sqrt{x}-2-2\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\frac{\sqrt{x}+2}{6}=\frac{-1}{\sqrt{x}-2}\)

b: D>0

=>\(\frac{-1}{\sqrt{x}-2}>0\)

=>\(\sqrt{x}-2<0\)

=>\(\sqrt{x}<2\)

=>0<=x<4

Bài 5:

\(C=\frac{\left(\sqrt{x}-\sqrt{y}\right)^2+4\sqrt{xy}}{\sqrt{x}+\sqrt{y}}-\frac{x\sqrt{y}-y\cdot\sqrt{x}}{\sqrt{xy}}\)

\(=\frac{x-2\sqrt{xy}+y+4\sqrt{xy}}{\sqrt{x}+\sqrt{y}}-\frac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\)

\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)=\sqrt{x}+\sqrt{y}-\left(\sqrt{x}-\sqrt{y}\right)=2\sqrt{y}\)

Bài 4:

a: ĐKXĐ: x>=1 và x<>3

b: \(B=\frac{x-3}{\sqrt{x-1}-\sqrt2}\)

\(=\frac{x-1-2}{\sqrt{x-1}-\sqrt2}\)

\(=\frac{\left(\sqrt{x-1}+\sqrt2\right)\left(\sqrt{x-1}-\sqrt2\right)}{\sqrt{x-1}-\sqrt2}=\sqrt{x-1}+\sqrt2\)

Thay \(x=4\left(2-\sqrt3\right)\) vào B, ta được:

\(B=\sqrt{4\left(2-\sqrt3\right)-1}+\sqrt2\)

\(=\sqrt{7-4\sqrt3}+\sqrt2=2-\sqrt3+\sqrt2\)

c: \(\sqrt{x-1}\ge0\forall x\) thỏa mãn ĐKXĐ
=>\(\sqrt{x-1}+\sqrt2\ge\sqrt2\forall x\) thỏa mãn ĐKXĐ

Dấu '=' xảy ra khi x-1=0

=>x=1

a, <=> 2,5 : 4x = 2,5

<=> 4x = 2,5 : 2,5 = 1

<=> x=1 : 4 = 1/4

b, <=> 1/5.x:3 = 8/3

<=> 1/5.x = 8/3 . 3 = 8

<=> x = 8 : 1/5 = 40

cai lon me may

ukm

ok !

câu nào vậy bợn

\(\frac{x^3-x^2-x-2}{x^5-3x^4+4x^3-5x^2+3x-2}\)

\(=\frac{x^3-2x^2+x^2-2x+x-2}{x^5-2x^4-x^4+2x^3+2x^3-4x^2-x^2+2x+x-2}\)

\(=\frac{\left(x^3-2x^2\right)+\left(x^2-2x\right)+\left(x-2\right)}{\left(x^5-2x^4\right)-\left(x^4-2x^3\right)+\left(2x^3-4x^2\right)-\left(x^2-2x\right)+\left(x-2\right)}\)

\(=\frac{x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)}{x^4\left(x-2\right)-x^3\left(x-2\right)+2x^2\left(x-2\right)-x\left(x-2\right)+\left(x-2\right)}\)

\(=\frac{\left(x-2\right)\left(x^2+x+1\right)}{\left(x-2\right)\left(x^4-x^3+2x^2-x+1\right)}=\frac{x^2+x+1}{x^4-x^3+2x^2-x+1}\)