(1+x)+(2+x)+(4+x)+(7+x)+.......+(22+x)=623
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1) Ta có: \(\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{7}{25}\cdot\dfrac{5}{7}\right)\)
\(=\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{1}{5}\right)\)
=0
2) Ta có: \(\dfrac{8}{17}\cdot\dfrac{4}{15}+\dfrac{8}{17}\cdot\dfrac{22}{15}-\dfrac{8}{15}\cdot\dfrac{9}{17}\)
\(=\dfrac{8}{17}\left(\dfrac{4}{15}+\dfrac{22}{15}-\dfrac{9}{15}\right)\)
\(=\dfrac{8}{17}\cdot\dfrac{15}{15}=\dfrac{8}{17}\)
3) Ta có: \(\dfrac{2021}{2}\cdot\dfrac{1}{3}+\dfrac{4042}{4}\cdot\dfrac{1}{5}+\dfrac{6063}{3}\cdot\dfrac{22}{15}\)
\(=\dfrac{2021}{2}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)+2021\cdot\dfrac{22}{15}\)
\(=\dfrac{2021}{2}\cdot\dfrac{8}{15}+\dfrac{2021}{2}\cdot\dfrac{44}{15}\)
\(=\dfrac{2021}{2}\cdot\dfrac{52}{15}\)
\(=\dfrac{52546}{15}\)
4) Ta có: \(\dfrac{4}{7}\cdot\dfrac{2}{13}+\dfrac{8}{13}:\dfrac{7}{4}+\dfrac{4}{7}:\dfrac{13}{2}+\dfrac{4}{7}\cdot\dfrac{1}{13}\)
\(=\dfrac{4}{7}\left(\dfrac{2}{13}+\dfrac{8}{13}+\dfrac{2}{13}+\dfrac{1}{13}\right)\)
\(=\dfrac{4}{7}\)
1, =4/9 x (3/7 + 2 x 4/7) - 1/9 - 3/9
= 4/9 x 1 - 1/9 - 3/9 = 0
2, = ( 22/5 x 5/22 ) x 12
= 1 x 12 = 12
1) 2(x + 5) + 3(x + 7) = 41
2x + 10 + 3x + 21 = 41
5x + 31 = 41
5x = 10
x = 2
6) 7(x - 1) + 5(3 - x) = 11x - 10
7x - 7 + 15 - 5x = 11x - 10
2x + 8 = 11x - 10
-9x = -18
x = 2
2) 5(x + 6) + 2(x - 3) = 38
5x + 30 + 2x - 6 = 38
7x + 24 = 38
7x = 14
x = 2
7) 4(2 + x) + 3(x - 2) = 12
8 + 4x + 3x - 6 = 12
7x + 2 = 12
7x = 10
x = 10/7
3) 7(5 + x) + 2(x - 10) = 15
35 + 7x + 2x - 20 = 15
9x + 15 = 15
9x = 0
x = 0
8) 5(2 + x) + 4(3 - x) = 10x - 15
10 + 5x + 12 - 4x = 10x - 15
x + 22 = 10x - 15
9x = 37
x = 37/9
4) 3(x + 4) + (8 - 2x) = 22
3x + 12 + 8 - 2x = 22
x + 20 = 22
x = 2
9) 7(x - 2) + 5(3 - x) = 11x - 6
7x - 14 + 15 - 5x = 11x - 6
2x + 1 = 11x - 6
-9x = -7
x = 7/9
5) 4(x + 5) + 3(7 - x) = 49
4x + 20 + 21 - 3x = 49
x + 41 = 49
x = 8
10) 5(3 - x) + 5(x + 4) = 6 + 4x
15 - 5x + 5x + 20 = 6 + 4x
35 = 6 + 4x
4x = 29
x = 29/4
a) \(\frac{22}{7}:\left(11-x\right)+\frac{2}{3}=\frac{7}{5}\)
\(\frac{22}{7}:\left(11-x\right)=\frac{7}{5}-\frac{2}{3}\)
\(\frac{22}{7}:\left(11-x\right)=\frac{11}{15}\)
\(11-x=\frac{22}{7}:\frac{11}{15}\)
\(11-x=\frac{30}{7}\)
\(x=11-\frac{30}{7}\)
\(x=\frac{47}{7}\)
b) \(22.\left(x+1\right)-12,5=53,5\)
\(22.\left(x+1\right)=53,5+12,5\)
\(22.\left(x+1\right)=66\)
\(x+1=66:22\)
\(x+1=3\)
\(x=3-1\)
\(x=2\)
c) \(\left(x-\frac{1}{2}\right).\frac{5}{2}+\frac{1}{2}=\frac{7}{4}\)
\(\left(x-\frac{1}{2}\right).\frac{5}{2}=\frac{7}{4}-\frac{1}{2}\)
\(\left(x-\frac{1}{2}\right).\frac{5}{2}=\frac{5}{4}\)
\(x-\frac{1}{2}=\frac{5}{4}:\frac{5}{2}\)
\(x-\frac{1}{2}=\frac{1}{2}\)
\(x=\frac{1}{2}+\frac{1}{2}\)
\(x=1\)
a)22/7:(11-x)=7/5-2/3
22/7:(11-x)=11/15
11-x=22/7:11/15
11-x=30/7
x=11-30/7
x=47/7
vậy x=47/7
b)22.(x+1)=53,5+12,5
22.(x+1)=66
x+1=66:2
x+1=3
x=3-1
x=2
vậy x=2
c)(x-1/2).5/2=7/4-1/2
(x-1/2).5/2=5/4
x-1/2=5/4:5/2
x-1/2=1/2
x=1/2+1/2
x=1
vậy x=1
bạn hãy tích đúng cho mình nhé
Bài 2:
Ta có: \(x\left(x-4\right)-x^2+8=0\)
\(\Leftrightarrow x^2-4x-x^2+8=0\)
\(\Leftrightarrow-4x=-8\)
hay x=2
b) Ta có: \(\left(x^2-7\right)\left(x+2\right)-\left(2x-1\right)\left(x-14\right)+x\left(x^2-2x-22\right)+35\)
\(=x^3+2x^2-7x-14-\left(2x^2-28x-x+14\right)+x^3-2x^2-22x+35\)
\(=2x^3-29x+21-2x^2+29x-14\)
\(=2x^3-2x^2+7\)
\(a,2\left(x-1\right)+3=x+2\)
\(\Leftrightarrow2x-2+3=x+2\)
\(\Leftrightarrow2x-x=2+2-3\)
\(\Leftrightarrow x=1\)
Vậy \(S=\left\{1\right\}\)
\(b,\left(3x-7\right)\left(x+5\right)=\left(5+x\right)\left(3-2x\right)\)
\(\Leftrightarrow\left(3x-7\right)\left(x+5\right)-\left(5+x\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(3x-7-3+2x\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(5x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\5x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
Vậy \(S=\left\{-5;2\right\}\)
=>1+x+2+x+4+x+7+x+11+x+16+x+22+x=623
=>(x+x+x+x+x+x+x)+(1+2+4+7+11+16+22)=623
=>7x+63=623
=>7x=560
=>x=560/7=80