79*50 = 3950

11+12+13+14+15+16+17+18+19+...+89+88+87+86+85+84+83+82

=( 11 + 89 )+(12+88)+(13+87)+(14+86)+(15+85)+(16+84)+(17+83)+(18+82)+19

=100+100+100+100+100+100+100+100+19

=800+19

=819

\(\dfrac{x+11}{89}+\dfrac{x+13}{87}=\dfrac{x+15}{85}+\dfrac{x+17}{83}\\\Leftrightarrow\left(\dfrac{x+11}{89}+1\right)+\left(\dfrac{x+13}{87}+1\right)=\left(\dfrac{x+15}{85}+1\right)+\left(\dfrac{x+17}{83}+1\right)\\ \Leftrightarrow\dfrac{x+100}{89}+\dfrac{x+100}{87}-\dfrac{x+100}{85}-\dfrac{x+100}{83}=0\\ \Leftrightarrow\left(x+100\right)\left(\dfrac{1}{89}+\dfrac{1}{87}-\dfrac{1}{85}-\dfrac{1}{83}\right)=0\\ \Leftrightarrow x+100=0\left(vì\dfrac{1}{89}+\dfrac{1}{87}-\dfrac{1}{85}-\dfrac{1}{83}\ne0\right)\\ \Leftrightarrow x=-100\)

\(\dfrac{x-11}{89}\) +\(\dfrac{x-13}{87}+\dfrac{x-15}{85}+\dfrac{x-17}{83}\)=4

⇔\(\dfrac{x-11}{89}+\dfrac{x-13}{87}+\dfrac{x-15}{85}+\dfrac{x-17}{83}-4=0\)

⇔\(\dfrac{x-11}{89}-1+\dfrac{x-13}{87}-1+\dfrac{x-15}{85}-1+\dfrac{x-17}{83}-1=0\)

⇔\(\dfrac{x-100}{89}+\dfrac{x-100}{87}+\dfrac{x-100}{85}+\dfrac{x-100}{83}=0\)

⇔\(\left(x-100\right)\left(\dfrac{1}{89}+\dfrac{1}{87}+\dfrac{1}{85}+\dfrac{1}{83}\right)=0\)

\(Do\) \(\dfrac{1}{89}+\dfrac{1}{87}+\dfrac{1}{85}+\dfrac{1}{83}\)≠\(0\) nên x-100=0 nên x=100

KL........

⇔(\(\dfrac{x-11}{89}\)-1)+(\(\dfrac{x-13}{87}\)-1)+(\(\dfrac{x-15}{85}\)-1)+(\(\dfrac{x-17}{83}\)-1)=0

⇔\(\dfrac{x-100}{89}\)+\(\dfrac{x-100}{87}\)+\(\dfrac{x-100}{85}\)+\(\dfrac{x-100}{83}\)=0

⇔(x-100)(\(\dfrac{1}{89}\)+\(\dfrac{1}{87}\)+\(\dfrac{1}{85}\)+\(\dfrac{1}{83}\))=0 (1)

Do 1/89+1/87+1/85+1/83≠0 nên (1)⇔x-100=0 ⇔x=100

Vậy tập nghiệm của PT là S=\(\left\{100\right\}\)

 $\frac{x-11}{89}+\frac{x-13}{87}+\frac{x-15}{85}+\frac{x-17}{83}=4$$=>\left(\frac{x-11}{89}-1\right)+\left(\frac{x-13}{87}-1\right)+\left(\frac{x-15}{85}-1\right)+\left(\frac{x-17}{83}-1\right)=0$$=>\frac{x-100}{89}+\frac{x-100}{87}+\frac{x-100}{85}+\frac{x-100}{83}=0$$=>\left(x-100\right)\left(\frac{1}{89}+\frac{1}{87}+\frac{1}{85}+\frac{1}{83}\right)=0$=> x-100 =0 => x=100Vậy nghiệm là 100 

ta có 0=\(\frac{x-11}{89}-1+\frac{x-13}{87}-1+\frac{x-15}{85}-1+\frac{x-17}{83}-1=0\)

0=\(\frac{x-11-89}{89}+\frac{x-13-87}{87}+\frac{x-15-85}{85}+\frac{x-17-83}{83}\)

0=\(\frac{x-100}{89}+\frac{x-100}{87}+\frac{x-100}{85}+\frac{x-100}{83}\)

0=(x-100)(\(\frac{1}{89}+\frac{1}{87}+\frac{1}{85}+\frac{1}{83}\))

vì \(\frac{1}{89}+\frac{1}{87}+\frac{1}{85}+\frac{1}{83}\)khác 0      suy ra x-100=0 

                                                                                    x=100

a, - 128 . (-25+89) + 128.(89-125)

   = 128 . - (-25 +89) +  128 .  (89 - 125)

   =  128 . [-(-25+89)+(89 - 125)]

   = 128. (25+89+89-125)

   = 128 . (89.2+-100)=128.78=(120+8).(70+8)= 8.190=1520

ai làm hộ mình câu b với

Số hạng tử của B là : ( 89 - 11 ) : 2 + 1 = 40 ( hạng tử )

\(B=11+13+15+...+87+89\)

\(=\left(11+98\right).40:2\)

\(=2180\)

B = 11 + 13 + 15 + ... + 87 + 89

Ta có : B = 11 + 13 + 15 + ... + 87 + 89 (có 40 số )

               = (89 + 11) . 40 : 2

               = 2000

Ta có:

$A=\frac{3^{10}.11+3^{10}.5}{3^9{.2}^4}=\frac{3^{10}.\left(11+5\right)}{3^9.16}$

$=\frac{3^{10}.16}{3^9.16}=\frac{3.1}{1.1}=3$

Vậy giá trị biểu thức A là 3

200 nheLouise Francoise

11 + 13 + 89 + 87

= ( 11 + 89 ) + ( 12 + 87 )

= 100 + 100

= 200