Bài toán được mô tả như hình sau:

Xét \(\Delta ABC\) vuông tại A có: 

\(\tan B=\dfrac{AC}{AB}\)

\(\Rightarrow AC=AB.\tan B=25.\tan40^{\circ}\approx21\left(m\right)=210\left(dm\right)\)

(x - 5) x 2 = 16

x - 5 = 16 : 2

x - 5 = 8

x = 8 + 5

x = 13

Vậy: ... 

\(\left(x-5\right)\times2=16\)

\(x-5=16:2\)

\(x-5=8\)

\(x=8+5\)

\(x=13\)

Vậy \(x=13\)

\(\left(2022\text{x}2023+2023\text{x}2024\right)\text{x}\left(2023\text{x}99-2022\text{x}99-99\right)\)

\(=2023\text{x}\left(2022+2024\right)\text{x}99\text{x}\left(2023-2022-1\right)\)

\(=0\text{x}2023\text{x}4046\text{x}99=0\)

(2022 x 2023 + 2023 x 2024) x (2023 x 99 - 2022 x 99 - 99)

= (2022 + 2024) x 2023 x (2023 - 2022 - 1) x 99

= (2022 + 2024) x 2023 x 0 x 99

= 0

Số khoai còn lại là:

\(25131\cdot\dfrac{2}{3}=16754\left(kg\right)\)

Số túi chia được là:

16754:3=5584(dư 2 kg)

cảm ơn

help

uh

Bài 2:

a: Số số hạng là \(999-1+1=999\left(số\right)\)

Tổng của dãy số là \(999\cdot\dfrac{\left(999+1\right)}{2}=999\cdot500=499500\)

b: Số số hạng là \(\dfrac{2010-10}{2}+1=\dfrac{2000}{2}+1=1001\left(số\right)\)

Tổng của dãy số là \(\left(2010+10\right)\times\dfrac{1001}{2}=1011010\)

c: Số số hạng là \(\dfrac{79-1}{3}+1=27\left(số\right)\)

Tổng của dãy số là: \(\left(79+1\right)\times\dfrac{27}{2}=27\times40=1080\)

d: Số số hạng là \(\dfrac{155-15}{2}+1=71\left(số\right)\)

Tổng của dãy số là (155+15)x71/2=6035

Bài 3:

a: 71-(33+x)=26

=>33+x=71-26=45

=>x=45-33=12

b: (x+73)-26=76

=>x+73-26=76

=>x+47=76

=>x=76-47=29

c: 45-(x+9)=6

=>x+9=45-6=39

=>x=39-9=30

d: 89-(73-x)=20

=>73-x=89-20=69

=>x=73-69=4

e: (x+7)-25=13

=>x+7=38

=>x=38-7=31

f: 198-(x+4)=120

=>x+4=198-120=78

=>x=78-4=74

m: 7x-5=16

=>7x=21

=>x=21:7=3

n: 156-2x=82

=>2x=156-82=74

=>x=74:2=37

k: 10x+65=125
=>10x=125-65=60

=>x=60:10=6

Bài 1:

a: 36+59+64

=(36+64)+59

=100+59=159

b: 123-79+77

=(123+77)-79

=200-79=121

c: \(23\cdot46+23\cdot64\)

\(=23\cdot\left(46+64\right)\)

\(=23\cdot110=2530\)

d: \(19\cdot34+19\cdot66-900\)

\(=19\cdot\left(34+66\right)-900\)

=1900-900=1000

e: \(58\cdot75+58\cdot50-58\cdot25\)

\(=58\left(75+50-25\right)\)

\(=58\cdot100=5800\)

f: \(27\cdot39+27\cdot63-2\cdot27=27\cdot\left(39+63-2\right)\)

\(=27\cdot100=2700\)

m: \(128\cdot46+128\cdot32+128\cdot22\)

\(=128\cdot\left(46+32+22\right)=128\cdot100=12800\)

n: \(66\cdot25+5\cdot66+66\cdot14+33\cdot66\)

\(=66\left(25+5+14+33\right)=66\cdot77=5082\)

k: \(12\cdot35+35\cdot182-35\cdot94\)

\(=35\cdot\left(12+182-94\right)\)

\(=35\cdot100=3500\)

h: \(35\cdot23+35\cdot41+64\cdot65\)

\(=35\cdot\left(23+41\right)+64\cdot65\)

\(=35\cdot64+64\cdot65=64\cdot100=6400\)

\(\dfrac{3}{2\text{x}4}+\dfrac{3}{4\text{x}6}+...+\dfrac{3}{48\text{x}50}\)

\(=\dfrac{3}{2}\text{x}\left(\dfrac{2}{2\text{x}4}+\dfrac{2}{4\text{x}6}+...+\dfrac{2}{48\text{x}50}\right)\)

\(=\dfrac{3}{2}\text{x}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{48}-\dfrac{1}{50}\right)\)

\(=\dfrac{3}{2}\text{x}\left(\dfrac{1}{2}-\dfrac{1}{50}\right)=\dfrac{3}{2}\text{x}\dfrac{24}{50}=\dfrac{36}{50}=\dfrac{18}{25}\)

\(\dfrac{2}{3}A=\dfrac{2}{2\times4}+\dfrac{2}{4\times6}+...\dfrac{2}{48\times50}\)

\(\dfrac{2}{3}A=\dfrac{4-2}{2\times4}+\dfrac{6-4}{4\times6}+...+\dfrac{50-48}{48\times50}\)

\(\dfrac{2}{3}A=\dfrac{4}{2\times4}-\dfrac{2}{2\times4}+\dfrac{6}{4\times6}-\dfrac{4}{4\times6}+...+\dfrac{50}{48\times50}-\dfrac{48}{48\times50}\)

\(\dfrac{2}{3}A=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{48}-\dfrac{1}{50}\)

\(\dfrac{2}{3}A=\dfrac{1}{2}-\dfrac{1}{50}\)

\(\dfrac{2}{3}A=\dfrac{12}{25}\)

\(A=\dfrac{12}{25}\div\dfrac{2}{3}\)

\(A=\dfrac{18}{25}\)

x+(x+1)+(x+2)+...+(x+10)=88

=>11x+(1+2+...+10)=88

=>11x+55=88

=>x+5=8

=>x=3

Bài 1:

\(\dfrac{x-2}{5}=\dfrac{-2}{2y+1}\)

=>\(\left(x-2\right)\left(2y+1\right)=5\cdot\left(-2\right)=-10\)

mà 2y+1 lẻ

nên \(\left(x-2;2y+1\right)\in\left\{\left(2;-5\right);\left(-2;5\right);\left(-10;1\right);\left(10;-1\right)\right\}\)

=>\(\left(x;y\right)\in\left\{\left(4;-3\right);\left(0;2\right);\left(-8;0\right);\left(12;-1\right)\right\}\)

Bài 2:

a: \(\left(2x-1\right)^2+4>=4\forall x\)

=>\(B=\dfrac{20}{\left(2x-1\right)^2+4}< =\dfrac{20}{4}=5\forall x\)

Dấu '=' xảy ra khi 2x-1=0

=>\(x=\dfrac{1}{2}\)

b: \(\left(x^2+1\right)^2>=1\forall x\)

=>\(\left(x^2+1\right)^2+5>=1+5=6\forall x\)

=>\(C=\dfrac{10}{\left(x^2+1\right)^2+5}< =\dfrac{10}{6}=\dfrac{5}{3}\forall x\)

Dấu '=' xảy ra khi x=0

BÀI 4A

\(\dfrac{1}{-2}+\dfrac{1}{-6}+\dfrac{1}{-12}+\dfrac{1}{-20}+\dfrac{1}{-30}+\dfrac{1}{-42}+\dfrac{1}{-56}+\dfrac{1}{-72}+\dfrac{1}{-90}\\ =-1\cdot\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\right)\\ =-1\cdot\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)\\ =-1\cdot\left(\dfrac{1}{1}-\dfrac{1}{10}\right)=-1\cdot\dfrac{9}{10}=-\dfrac{9}{10}\)