a. \(C=\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-\frac{1}{21}+...+\frac{1}{61}-\frac{1}{66}\)

\(=\frac{1}{11}-\frac{1}{66}=\frac{5}{66}\)

b. \(D=\frac{2}{3}.\left(\frac{3}{1.4}+\frac{4}{4.7}+...+\frac{3}{97.100}\right)\)

\(=\frac{2}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{100}\right)\)

\(=\frac{2}{3}.\left(1-\frac{1}{100}\right)=\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)

\(C=\frac{1}{11}-\frac{1}{16}+\frac{1}{16}-....-\frac{1}{66}\)

\(C=\frac{1}{11}-\frac{1}{66}=\frac{5}{66}\)

\(D=\frac{2}{3}.\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-....-\frac{1}{100}\right)\)

\(D=\frac{2}{3}.\left(1-\frac{1}{100}\right)=\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)

nêu viết đề bài như thế thì nghĩ như là\(\frac{2\cdot4}{1}+\frac{2\cdot7}{4}...\)
=> hai bạn kia sai

\(\left(\dfrac{6:\dfrac{3}{5}-\dfrac{17}{16}.\dfrac{6}{7}}{\dfrac{21}{5}.\dfrac{10}{11}+\dfrac{57}{11}}-\dfrac{\left(\dfrac{3}{20}+\dfrac{1}{2}-\dfrac{1}{15}\right).\dfrac{12}{49}}{\dfrac{10}{3}+\dfrac{2}{9}}\right).x=\dfrac{215}{96}\)

\(\Rightarrow\left(\dfrac{\dfrac{509}{56}}{9}-\dfrac{\dfrac{7}{12}.\dfrac{12}{49}}{\dfrac{32}{9}}\right).x=\dfrac{215}{96}\)

\(\Rightarrow\left(\dfrac{509}{504}-\dfrac{\dfrac{1}{7}}{\dfrac{32}{9}}\right).x=\dfrac{215}{96}\)

\(\Rightarrow\left(\dfrac{509}{504}-\dfrac{9}{224}\right).x=\dfrac{215}{96}\)

\(\Rightarrow\dfrac{1955}{2016}.x=\dfrac{215}{96}\)

\(\Rightarrow x=\dfrac{215}{96}:\dfrac{1955}{2016}\)

\(\Rightarrow x=\dfrac{903}{391}\)

`[ 6 : 3/5 - 17/16 . 6/7 : 21/5 . 10/11 + 57/11 -  (3/20 + 1/2 - 1/15) . 12/49 :  10/3 + 2/9 ] . x = 215/96`

`=>[ 6  . 5/3 - 17/16 . 6/7 . 5/21 . 10/11 + 57/11 -  (3/20 + 1/2 - 1/15) . 12/49 . 3/10 + 2/9 ] . x = 215/96`

`=>[10- 51/56 . 6/7 . 5/21 . 10/11 + 57/11 -  (3/20 + 1/2 - 1/15) . 12/49 . 3/10 + 2/9 ] . x = 215/96`

`=> [10- 153/196  . 5/21 . 10/11 + 57/11 -  (3/20 + 1/2 - 1/15) . 12/49 . 3/10 + 2/9 ] . x = 215/96`

`=> [10- 255/1372 . 10/11 + 57/11 -  (3/20 + 1/2 - 1/15) . 12/49 . 3/10 + 2/9 ] . x = 215/96`

`=> [10- 1275/7546 + 57/11 -  (3/20 + 1/2 - 1/15) . 12/49 . 3/10 + 2/9 ] . x = 215/96`

`=> (10- 1275/7546 + 57/11 -  7/12. 12/49 . 3/10 + 2/9 ) . x = 215/96`

`=> ( 10- 1275/7546 + 57/11 -343/600 . 3/10 + 2/9 ) . x = 215/96`

`=> ( 10- 1275/7546 + 57/11 -343/2000 + 2/9 ) . x = 215/96`

`=>15,06357671 . x= 215/96`

`=> x= 215/96: 15,06357671`

`=>x= 0,1486754027`

Đề có phải như vậy không nhỉ ?

Bạn dùng công thức này mà làm nhé :\(\frac{n}{a\times\left(a+n\right)}=\frac{1}{a}-\frac{1}{a+n}\)

Ví dụ :\(\frac{2}{3\times5}=\frac{1}{3}-\frac{1}{5};\frac{3}{4\times7}=\frac{1}{4}-\frac{1}{7};\frac{5}{6\times11}=\frac{1}{6}-\frac{1}{11}\)

Bài 46:

11: Ta có: \(-4\left|x-2\right|=-8\)

\(\Leftrightarrow\left|x-2\right|=2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=2\\x-2=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\)

Vậy: x∈{0;4}

12: Ta có: \(5\left|x+2\right|=-10\cdot\left(-2\right)\)

\(\Leftrightarrow5\left|x+2\right|=20\)

\(\Leftrightarrow\left|x+2\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)

Vậy: x∈{-6;2}

13: Ta có: \(6\left|x-2\right|=18:\left(-3\right)\)

\(\Leftrightarrow6\left|x-2\right|=-6\)(1)

Ta có: \(\left|x-2\right|\ge0\forall x\)

\(\Rightarrow6\left|x-2\right|\ge0\forall x\)(2)

Ta có: -6<0(3)

Từ (1), (2) và (3) suy ra x∈∅

Vậy: x∈∅

14: Ta có:\(-7\left|x+4\right|=21:\left(-3\right)\)

\(\Leftrightarrow-7\left|x+4\right|=-7\)

\(\Leftrightarrow\left|x+4\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=1\\x+4=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)

Vậy: x∈{-5;-3}

15: Ta có: \(4\left|x+1\right|=8\left(-2\right)-8\left(-5\right)\)

\(\Leftrightarrow4\left|x+1\right|=-16-\left(-40\right)\)

\(\Leftrightarrow4\left|x+1\right|=24\)

\(\Leftrightarrow\left|x+1\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=6\\x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-7\end{matrix}\right.\)

Vậy: x∈{-7;5}

16: Ta có: \(3\left|x+5\right|=-9\)(4)

Ta có: |x+5|≥0∀x

⇒3|x+5|≥0∀x(5)

Ta có: -9<0(6)

Từ (4), (5) và (6) suy ra x∈∅

Vậy: x∈∅

17: Ta có: \(-8\left|x-3\right|=24-16:2\)

\(\Leftrightarrow-8\left|x-3\right|=16\)

\(\Leftrightarrow\left|x-3\right|=-2\)

mà |x-3|≥0>-2∀x

nên x∈∅

Vậy: x∈∅

18: Ta có: \(-3\left|x+6\right|=6\cdot2-9\)

\(\Leftrightarrow-3\left|x+6\right|=3\)

\(\Leftrightarrow\left|x+6\right|=-1\)

mà |x+6|≥0>-1∀x

nên x∈∅

Vậy: x∈∅

19: Ta có: \(5-\left|x+7\right|=4\)

\(\Leftrightarrow\left|x+7\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x+7=-1\\x+7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=-6\end{matrix}\right.\)

Vậy: x∈{-8;-6}

20: Ta có: \(12-\left|x+8\right|=10\)

\(\Leftrightarrow\left|x+8\right|=2\)

\(\Leftrightarrow\left[{}\begin{matrix}x+8=2\\x+8=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=-10\end{matrix}\right.\)

Vậy: x∈{-10;-6}

lớp 1 mà cậu

4.24.5²-(3³.18+3³.12)

=4.24.25-[27.(18+12)]

=(4.25).24-[27.30]

=100.24-810

=2400-810

=1590

4.24.5^2-(3^3.18+3^3.12)

31.15.7^2.4-31.49.40

1+2+3+4+5+6+7+8+9+10

1+3+5+7+9+11+13+15+19

2+6+10+14+22+23+26+34

5+8+11+14+17+20+23+26+29

1+6+11+16+21+26+31+36+41+47+51

10+13+16+19+22+25+28+31+34+37+40

5+7+9+11+13+15+17+3+8+13+18+23+28

4+7+10+13+16+19+5+9+13+17+21+25 = 1590

#HT#

1) x - 2 = -6

x = -6 + 2

x = -4

2) -5 . x - ( -3 ) =13

-5 . x = 13 + ( -3 )

-5 . x = 10

x = 10 : ( -5 )

x = -2

98775 - 32 x 85

=98775 -2720

=96055

 67500 - 24 x 236

= 67500 -5664

=61836

 568 + 101598 : 287

= 568 +354

=922

6875 + 980 -180  

=7855 -180 

=7675

\(\dfrac{2}{5}+\dfrac{3}{10}-\dfrac{1}{2}\)

\(=\dfrac{7}{10}-\dfrac{1}{2}\)

= \(\dfrac{1}{5}\)

\(\dfrac{8}{11}+\dfrac{8}{33}x\dfrac{3}{4}\)

\(=\dfrac{8}{11}+\dfrac{2}{11}\)

\(=\dfrac{10}{11}\)

\(\dfrac{7}{9}x\dfrac{3}{14}:\dfrac{5}{8}\)

\(=\dfrac{1}{6}:\dfrac{5}{8}\)

\(=\dfrac{1}{6}x\dfrac{8}{5}\)

\(=\dfrac{8}{30}\)

\(=\dfrac{4}{15}\)

\(\dfrac{5}{12}-\dfrac{7}{32}:\dfrac{21}{16}\)

\(=\dfrac{5}{12}-\dfrac{7}{32}x\dfrac{16}{21}\)

\(=\dfrac{5}{12}-\dfrac{1}{6}\)

\(=\dfrac{5}{12}-\dfrac{2}{12}\)

\(=\dfrac{3}{12}=\dfrac{1}{4}\)

_{Tín hịu cầu kíu :>>}

20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)

Vậy...

1) 16 - 8x = 0 ⇔ 8(2 - x) = 0⇔ 2 - x = 0 ⇔ x = 2

Vậy phương trình có nghiệm là x = 2