Lớp 6A góp được 200*1/5=40 quyển

Lớp 6B góp được 40*150%=60 quyển

Lớp 6C góp được:

200-40-60=100 quyển

\(B=\dfrac{1}{6.10}+\dfrac{1}{10.14}+\dfrac{1}{14.18}+...+\dfrac{1}{402.406}\)

\(4B=\dfrac{4}{6.10}+\dfrac{4}{10.14}+\dfrac{4}{14.18}...+\dfrac{4}{402.406}\)

\(=\dfrac{1}{6}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{18}+...+\dfrac{1}{398}-\dfrac{1}{402}+\dfrac{1}{402}-\dfrac{1}{406}\)

\(=\dfrac{1}{6}-\dfrac{1}{406}=\dfrac{100}{609}\)

\(\Rightarrow B=\dfrac{25}{609}\)

\(B=\dfrac{1}{4}\cdot\left(\dfrac{4}{6\cdot10}+\dfrac{4}{10\cdot14}+...+\dfrac{4}{402\cdot406}\right)\)

=1/4(1/6-1/10+1/10-1/14+...+1/402-1/406)

=1/4*(1/6-1/406)

=1/4*400/(6*406)

=100/(6*406)=25/609

\(A=\dfrac{-1}{199}-\dfrac{1}{199\cdot198}-\dfrac{1}{198\cdot197}-...-\dfrac{1}{3\cdot2}-\dfrac{1}{2}\)

\(A=\dfrac{-1}{199}-\left(\dfrac{1}{198}-\dfrac{1}{199}\right)-\left(\dfrac{1}{197}-\dfrac{1}{198}\right)-...-\left(\dfrac{1}{2}-\dfrac{1}{3}\right)-\dfrac{1}{2}\)

\(A=\dfrac{-1}{199}-\dfrac{1}{198}+\dfrac{1}{199}-\dfrac{1}{197}+\dfrac{1}{198}-...-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{2}\)

\(A=-\dfrac{1}{2}-\dfrac{1}{2}\)

\(A=\dfrac{-1-1}{2}\)

\(A=\dfrac{-2}{2}\)

\(A=-1\)

\(=\dfrac{-1}{199}-\left(\dfrac{1}{2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{198\cdot199}\right)\)

\(=\dfrac{-1}{199}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{198}-\dfrac{1}{199}\right)\)

=-1/199-(1-1/199)

=-1/199-1+1/199=-1

-0.24>-4/5

1:

a:

góc DAB+góc CAE=180 độ-góc BAE=90 độ

góc DAB+góc DBA=90 độ

=>góc DBA=góc CAE

Xét ΔDBA vuông tại D và ΔEAC vuông tại E có

BA=AC
góc DBA=góc EAC

=>ΔDBA=ΔEAC

b: ΔDBA=ΔEAC

=>DB=EA và DA=EC

BD+CE

=CA+AD

=CD

a) \(\left(2x+5\right)\left(3x-1\right)-6x\left(x-3\right)-21x\)

\(=6x^2-2x+15x-5-6x^2+18x-21x\)

\(=\left(6x^2-6x^2\right)-\left(2x-15x-18x+21x\right)-5\)

\(=10x-5\)

b) \(\left(x+y\right)^2+\left(x-y\right)^2-2x^2\)

\(=x^2+2xy+y^2+x^2-2xy+y^2-2x^2\)

\(=\left(x^2+x^2-2x^2\right)+\left(2xy-2xy\right)+\left(y^2+y^2\right)\)

\(=2y^2\)

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

\(=\left[\left(x-1\right)-\left(x-3\right)\right]^2\)

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

\(=2^2\)

\(=4\)

Vậy: ...

b) \(\left(x-1\right)^3-\left(x+2\right)\left(x^2-2x+4\right)+3x^2-3x\)

\(=x^3-3x^2+3x-1-\left(x^3+8\right)+3x^2-3x\)

\(=\left(x^3-x^3\right)-\left(3x^2-3x^2\right)+\left(3x-3x\right)-\left(1+8\right)\)

\(=-9\)

Vậy: ...

e) \(\left(x-y+z\right)^2+\left(y-z\right)^2+2\left(x-y+z\right)\left(y-z\right)\)

\(=\left(x-y+z\right)^2+2\left(x-y+z\right)\left(y-z\right)+\left(y-z\right)^2\)

\(=\left[\left(x-y+z\right)+\left(y-z\right)\right]^2\)

\(=\left(x-y+z+y-z\right)^2\)

\(=x^2\)

f) \(\left(5x+1\right)^2-\left(5x+3\right)\left(5x-3\right)\)

\(=25x^2+10x+1-\left(25x^2-9\right)\)

\(=25x^2+10x+1-25x^2+9\)

\(=10x+10\)

a: =6x^2-2x+15x-5-6x^2+18x-21x

=13x+18x-21x-5

=10x-5

b; =x^2+2xy+y^2+x^2-2xy+y^2-2x^2

=2x^2+2y^2-2x^2

=2y^2

c: =9x^2+24x+16+16x^2-8x+1+4x^2-25

=29x^2+16x-8

d: =8x^3+1+8-27x^3

=-19x^3+9

Ta có:

\(P\left(x\right)=-4x^4+8x^2-12x+5\)

\(P\left(x\right)=-4x\left(x^3-2x+3\right)+5\)

Thay \(x^3-2x+3=0\) vào P(x) ta có:

\(P\left(x\right)=-4x\cdot0+5=0=5\)

⇒ Chọn B

P=-4x(x^3-2x+3)+5

=-4x*0+5=5

=>B