Bài 3: 

= 1- 1/2 + 1/2 -1/3 +...+ 1/98 -1/99

= 1- 1/99

= 98/99

Bài 4:

= 1/2*3 + 1/3*4 + 1/4*5 +...+  1/10*11

= 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 +...+ 1/10 - 1/11

= 1/2 - 1/11= 9/22

Bài 2:

a) \(=x^2-36y^2\)

b) \(=x^3-8\)

Bài 3:

a) \(=x^2+2x+1-x^2+2x-1-3x^2+3=-3x^2+4x+3\)

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

bài 1:

a) (x+1)^2-(x-1)^2-3(x+1)(x-1)

=(x+1+x-1)(x+1-x+1)-3x^2-3

=2x^2-3x^2-3

=-x^2-3

Bài 1 : 

\(\left|2x-1\right|=x-1\)ĐK : \(x\ge1\)

TH1 : \(2x-1=x-1\Leftrightarrow x=0\)(ktm)

TH2 : \(2x-1=1-x\Leftrightarrow3x=2\Leftrightarrow x=-\frac{2}{3}\)(ktm)

Vậy biểu thức ko có x thỏa mãn 

Bài 2 : 

\(\left|3x-1\right|=2x+3\)ĐK : x >= -3/2 

TH1 : \(3x-1=2x+3\Leftrightarrow x=4\)

TH2 : \(3x-1=-2x-3\Leftrightarrow5x=-2\Leftrightarrow x=-\frac{2}{5}\)

Nếu ol thì tham khảo nah nguoiemtinhthong.

1.1

2x2+5x−1=7x3−1−−−−−√2x2+5x−1=7x3−1

⇔2(x2+x+1)+3(x−1)−7(x−1)(x2+x+1)−−−−−−−−−−−−−−−√(1)⇔2(x2+x+1)+3(x−1)−7(x−1)(x2+x+1)(1)

Đặt a=x−1−−−−−√;b=x2+x+1−−−−−−−−√;a≥0;b>0a=x−1;b=x2+x+1;a≥0;b>0

pt (1) trở thành 3a2+2b2−7ab=03a2+2b2−7ab=0

a=2ba=2b v a=13ba=13b

Các bạn tự giải quyết tiếp nhé.

1.2

TXĐ D=[1;+∞)D=[1;+∞)

đặt a=x−1−−−−−√4;b=x+1−−−−−√4;a,b≥0a=x−14;b=x+14;a,b≥0

pt (2) trở thành 3a2+2b2−5ab=03a2+2b2−5ab=0

⇔a=b⇔a=b v a=23ba=23b

...

1.3

D=[3;+∞)D=[3;+∞)

Đặt a=x2+4x−5−−−−−−−−−√;b=x−3−−−−−√;a,b≥0a=x2+4x−5;b=x−3;a,b≥0

pt (3) trở thành 3a+b=11a2−19b2−−−−−−−−−√3a+b=11a2−19b2

⇔2a2−6ab−20b2=0⇔2a2−6ab−20b2=0

⇒a=5b⇒a=5b
...

1.4

ĐK

⇔2x2−2x+2=3(x−2)x(x+1)−−−−−−−−−−−−√2x2−2x+2=3(x−2)x(x+1)

⇔(x2−2x)+2(x+1)=3(x2−2x)(x+1)−−−−−−−−−−−−−√2(x2−2x)+2(x+1)=3(x2−2x)(x+1)

Đặt x2−2x−−−−−−√=ax2−2x=a; x+1−−−−−√=bx+1=b (a;b\geq0)

⇔2a2+2b2=3ab

1.5

Đặt 4x2−4x−10=t4x2−4x−10=t (t \geq 0)

⇔t=t+4x2−2x−−−−−−−−−−√t=t+4x2−2x

⇔t2−t−4x2+2x=0t2−t−4x2+2x=0

Δ=1−4(2x−4x2)=(4x−1)2Δ=1−4(2x−4x2)=(4x−1)2

⇒t=1−2xt=1−2x hoặc t=2xt=2x

1.1

2.2+5.-1=7.3-1-----v2.2+5.-1=7.3-1

2(.2+x+1)+3(x-1)

3a+b=11a2-19b2

tóm tắt

\(\frac{1}{2}\)x \(\frac{2}{3}\)x \(\frac{3}{4}\)x \(\frac{4}{5}\)x \(\frac{5}{6}\)=  \(\frac{1}{6}\)

Giải thích : Khử 2 với 2 ; 3 với 3 ; ... Cuối cùng còn lại 1 ở tử số và 6 ở mẫu số

1) 1/1.2 + 1/2.3 + ... + 1/6.7

= 1 - 1/2 + 1/2 - 1/3 + ... + 1/6 - 1/7

= 1 - 1/7

= 6/7

2) 1/2 + 1/6 + 1/12 + .. + 1/72

= 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/8.9

= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/8 - 1/9

= 1 - 1/9

= 8/9

3) \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2019}\right)\)

= \(\frac{1}{2}.\frac{2}{3}...\frac{2019}{2020}\)

= \(\frac{1.2....2019}{2.3...2020}\)

= \(\frac{1}{2020}\)

4) A = \(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{512}\)

       = \(\frac{1}{2^2}+\frac{2}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^9}\)

=> 2A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}\)

Lấy 2A - A = \(\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^8}\right)-\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^9}\right)\)

             A  = \(\frac{1}{2}-\frac{1}{2^9}\)