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8.6+ 288: ( x -3)\(^2\)= 503

48+ 288: (x-3)\(^2\)= 50

      288: (x-3)\(^2\)= 50-48

      288: (x-3)\(^2\)= 2

              (x-3)\(^2\)= 288:2

              (x-3)\(^2\)= 144

              (x-3)\(^2\)= 12\(^2\)

=>x- 3= 12=> x= 12+ 3x= 15

lm đúng tui tick cho 2 tick!

x= 56 tick tớ nhé 

\(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+99\right)=2021\)

\(\left(x+x+x+...+x\right)+\left(1+2+...+99\right)=2021\)

\(100x+\left(1+2+...+99\right)=2021\)

Ta tính tổng \(A=1+2+...+99\)  (Số số hạng của tổng là 99)

\(A=\left(1+99\right)+\left(2+98\right)+...+\left(49+51\right)+50\)

\(A=100+100+...+100+50=100\times49+50=4950\)

Vậy \(100x+4950=2021\)

Suy ra \(100x=2021-4950=-2929\), hay \(x=-29,29\)

Nguyễn Huy Thắng giải sai rồi ,thế này mới đúng nè

1,\(\frac{1}{6}+\frac{1}{12}+.........+\frac{1}{72}\)

=\(\frac{1}{2.3}+\frac{1}{3.4}+.......+\frac{1}{8.9}\)

=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.......+\frac{1}{8}-\frac{1}{9}\)

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

=\(\frac{7}{18}\)

2,\(\frac{3}{1.4}+\frac{3}{4.7}+..........+\frac{3}{13.16}\)

=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.........+\frac{1}{13}-\frac{1}{16}\)

=\(1-\frac{1}{16}\)

=\(\frac{15}{16}\)

2)đặt B= 3/1.4+3/4.7+3/7.10+3/10.13+3/13.16

\(B=3\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{16}\right)\)

\(B=3-\frac{15}{16}\)

\(B=\frac{45}{16}\)

\(\left(a^2+3b^2\right)\left(1+3\right)\ge\left(a+3b\right)^2\)

\(\Rightarrow\sqrt{a^2+3b^2}\ge\sqrt{\dfrac{\left(a+3b\right)^2}{4}}=\dfrac{a+3b}{2}\)

Tương tự:

\(\sqrt{b^2+3c^2}\ge\dfrac{b+3c}{2}\) ; \(\sqrt{c^2+3a^2}\ge\dfrac{c+3a}{2}\)

 Cộng vế \(\Rightarrow VT\ge\dfrac{4\left(a+b+c\right)}{2}=6\)

Dấu "=" xảy ra khi \(a=b=c=1\)

toán lớp 6 đây á

bằng 999 nha bạn

lộn sory

a, x.25=380125

x=380125:25

x=15205

b, (45,2+45.2)+2+x=2525255

45.2.2

45 . 2 + 2 + x + 45 . 2 = 2525255

=> 2 . (45 + 45 + 1) + x = 2525255

=> 182 + x = 2525255

=> x = 2525073

Coi \(A=\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}\)

\(\Rightarrow\frac{1}{2}A=\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+\frac{1}{45}\right).\frac{1}{2}\)

\(=\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}\)

\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\)

\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)

\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)

\(\Rightarrow A=\frac{2}{5}:\frac{1}{2}=\frac{4}{5}\)

Trần Thùy Dung: hay cho l.i.k.e

`Answer:`

a. \(\left(x-3\right)+\left(x-2\right)+\left(x-1\right)+...+9+10+11=11\)

\(\Leftrightarrow x-3+x-2+x-1+...+9+10+11=11\)

\(\Leftrightarrow3x+\left(-3-2-1+0+...+11\right)=11\)

\(\Leftrightarrow3x+60=11\)

\(\Leftrightarrow3x=-49\)

\(\Leftrightarrow x=-\frac{49}{3}\)

b. \(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=5750\)

\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(1+2+3+...+100\right)=5750\)

\(\Leftrightarrow100x+[\left(100+1\right).100:2]=5750\)

\(\Leftrightarrow100x+5050=5750\)

\(\Leftrightarrow100x=700\)

\(\Leftrightarrow x=7\)

c. \(2^x+2^{x+1}+2^{x+2}=960-2^{x+3}\)

\(\Leftrightarrow2^x+2^x.2+2^x.2^2+2^x.2^3=960\)

\(\Leftrightarrow2^x+2.2^x+4.2^x+8.2^x=960\)

\(\Leftrightarrow2^x.\left(1+2+4+8\right)=960\)

\(\Leftrightarrow2^x.15=960\)

\(\Leftrightarrow2^x=64\)

\(\Leftrightarrow x=6\)

Ta có 3xy+2x-5y=6

\(\Rightarrow\) 3xy+3x-x-5y=6

\(\Rightarrow\) 3x(y+1)-x-5y=6

\(\Rightarrow\) 3x(y+1)-x-5y-5=6-5

\(\Rightarrow\) 3x(y+1)-x-(5y+5)=1

\(\Rightarrow\) 3x(y+1)-x-5(y+1)=1

\(\Rightarrow\) (3x-5)(y+1)-x=1

\(\Rightarrow\) 3(3x-5)(y+1)-3x=3

\(\Rightarrow\) (3x-5)(3y+3)-3x+5=3+5

\(\Rightarrow\) (3x-5)(3y+3)-(3x-5)=8

\(\Rightarrow\) (3x-5)(3y+3-1)=8

\(\Rightarrow\) (3x-5)(3y+2)=8

Vì x,y nguyên nên 3x-5 nguyên, 3y+2 nguyên.

Mà (3x-5)(3y+2)=8 nên 3x-5 \(\in\)Ư(8), 3y+2\(\in\)Ư(8).

Ta có bảng sau:

Vậy các cặp số nguyên (x;y) là (