Tìm x
255-5.(x+3)=102
Tìm x:
255-5.(x+3)=102
5(x+3)=225-102
x+3=123/5
x=123/5-3
x=108/5
225-5(x+3)=102
=>225-5x-15=102
=>210-5x=102
=>5x=108
=>x=\(\frac{108}{5}\)
Bài 3:Tìm x∈N biết:
a) 70-5(x-3)=45 b) (2x-1)4 =3.62 -27
c) 3x3 +43=102 -33 d) 3x+2 + 3x =2430
e)24.x-32.x=145-255:51
Mn bày e gấp.Em đag cần gấp ạ
a) 70 - 5(x - 3 ) = 45
5( x - 3 ) = 70 - 45 = 25
x - 3 = 25 : 5 = 5
x = 5 + 3 = 8
b) (2x - 1 )4 = 3 . 62 - 27
(2x - 1 )4 = 3 . 36 - 27
(2x - 1 )4 = 81
Ta thấy 81 = 34 vậy suy ra (2x - 1)4 = 34
Để vế trong ngoặc tròn (2x - 1 ) = 3 thì x cần bằng 2
Thử lại : 2 . 2 - 1 = 4 - 1 = 3
Vậy x = 2
c) 3x3 + 43 = 102 - 33
3x3 + 43 = 100 - 33 = 67
3x3 = 67 + 43 = 110 ( Đoạn này đề bài sai hay tao sai z :)?)
Tìm X:
a, |5/3 -x| -|-5/6|=|-5/9|
b,|x+1/102|+|x+2/102|+...+|x+100/102| = 102x
a) |5/3 - x| - |-5/6| = |-5/9|
=> |5/3 - x| - 5/6 = 5/9
=> |5/3 - x| = 5/9 + 5/6
=> |5/3 - x| = 25/18
=> \(\orbr{\begin{cases}\frac{5}{3}-x=\frac{25}{18}\\\frac{5}{3}-x=-\frac{25}{18}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{5}{18}\\x=\frac{55}{18}\end{cases}}\)
a, \(\left|\frac{5}{3}-x\right|-\left|-\frac{5}{6}\right|=\left|-\frac{5}{9}\right|\)
\(\Leftrightarrow\left|\frac{5}{3}-x\right|-\frac{5}{6}=\frac{5}{9}\Rightarrow\left|\frac{5}{3}-x\right|=\frac{5}{9}+\frac{5}{6}=\frac{25}{18}\)
\(\Rightarrow\orbr{\begin{cases}\frac{5}{3}-x=\frac{25}{18}\\\frac{5}{3}-x=-\frac{25}{18}\end{cases}\Rightarrow}x.\)
b, \(\left|x+\frac{1}{102}\right|+\left|x+\frac{2}{102}\right|+...+\left|x+\frac{100}{102}\right|\ge0\)
\(\Rightarrow102x\ge0\Leftrightarrow x\ge0\)=> Ta có thể phá dấu GTTĐ
\(\Rightarrow x+\frac{1}{102}+x+\frac{2}{102}+...+x+\frac{100}{102}=102x\)
\(\Rightarrow100x+\frac{1+2+3+...+100}{102}=102x\Rightarrow2x\Rightarrow x.\)
Cho 1/x+1/y+1/z=0 (x; y; z khác 0). Chứng minh rằng: 1/x^2+1/y^2+1/z^2=3/xyz
\(\frac{1}{x}=a;\frac{1}{y}=b;\frac{1}{z}=c\left(x,y,z\ne0\right)\).
Ta có:
\(a+b+c=0\).
Ta phải chứng minh rằng nếu \(a+b+c=0\)thì \(a^3+b^3+c^3=3abc\).
Thật vậy, xét hiệu \(A=a^3+b^3+c^3-3abc\)với \(a+b+c=0\).
\(A=\left(a+b\right)^3-3ab\left(a+b\right)-3abc+c^3\).
\(A=\left[\left(a+b\right)^3+c^3\right]-\left[3ab\left(a+b\right)+3abc\right]\).
\(A=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]\)\(-3ab\left(a+b+c\right)\).
\(A=\left(a+b+c\right)\left(a^2+2ab+b^2-ab-ac+c^2-3ab\right)\).
\(A=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\).
\(A=0\left(a^2+b^2+c^2-ab-bc-ca\right)=0\)(vì \(a+b+c=0\)).
Do đó \(a^3+b^3+c^3-3abc=0\).
\(\Rightarrow a^3+b^3+c^3=3abc\)với \(a+b+c=0\)
\(\Rightarrow\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}\)với \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)(điều phải chứng minh).
(x + 129):4 = 844
(x - 452) x 5 = 255
(320 + x) - 102 =197
viết chi tiết
= 3184
= 503
K mk nhé , mà con cuối cùng kết quả là - 21 (âm 21) cơ chứ nếu thật thì ko trừ được đâu .( * . * )
(x + 129) : 4 = 844 (x - 452) x 5 = 255; (320 + x) - 102 = 197
x + 129 = 844 .4 x - 452 = 255 : 5 320 + x = 197 + 102
x + 129 = 3376 x - 452 = 51 320 + x = 299
x = 3376 - 129 x = 51 + 452 x = 320 - 299
x = 3247 x = 503 x = 21
(x+129):4= 844 (x-452) x 5 = 255 (320 + x) - 102 = 197
x+129 = 844.4 x-452 = 255:5 320 + x = 197 + 102
x+129 = 3376 x-452 = 51 320 + x = 299
x = 3376-129 x = 51+452 x = 299 - 320
x = 3247 x = 503 x = -21
102+(26-3.x):5=106
Tìm x
(26-3x):5=106-102=4
26-3x=4.5=20
3x=26-20=6
x=6:3
x=2
\(102+\left(26-3.x\right):5=106\)
\(\left(26-3x\right):5=106-102=4\)
\(26-3x=4.5=20\)
\(3x=26-20=6\)
\(x=6:3=2\)
tìm y biết : y+7/2 x y+ y x 3/5 = 102
`y+7/2xxy+yxx3/5=102`
`y+7/2xxyxx1xx+yxx3/5=102`
`yxx(7/2+1+3/5)=102`
`yxx5,1=102`
`y=102:5,1`
`y=20`
Vậy..
`@An`
Tìm x, biết:
\(a)\frac{x+1}{65}+\frac{x+2}{64}=\frac{x+3}{63}+\frac{x+4}{62}\)
\(b)\frac{x-5}{100}+\frac{x-4}{101}+\frac{x-3}{102}=\frac{x-100}{5}+\frac{x-101}{4}+\frac{x-102}{3}\)
\(a,\frac{x+1}{65}+\frac{x+2}{64}=\frac{x+3}{63}+\frac{x+4}{62}\)
\(\Rightarrow\left[\frac{x+1}{65}+1\right]+\left[\frac{x+2}{64}+1\right]=\left[\frac{x+3}{63}+1\right]+\left[\frac{x+4}{62}+1\right]\)
\(\Rightarrow\frac{x+1+65}{65}+\frac{x+2+64}{64}=\frac{x+3+63}{63}+\frac{x+4+62}{62}\)
\(\Rightarrow\frac{x+66}{65}+\frac{x+66}{64}=\frac{x+66}{63}+\frac{x+66}{62}\)
\(\Rightarrow\frac{x+66}{65}+\frac{x+66}{64}=\frac{x+66}{63}+\frac{x+66}{62}=0\)
\(\Rightarrow\left[x+66\right]\left[\frac{1}{65}+\frac{1}{64}-\frac{1}{63}+\frac{1}{62}\right]=0\)
Mà \(\frac{1}{65}+\frac{1}{64}-\frac{1}{63}+\frac{1}{62}\ne0\)
\(\Rightarrow x+66=0\)
\(\Rightarrow x=0-66=-66\)
Auto làm nốt câu b
a, Cộng cả 2 vế với 2
Ta có \(\frac{x+1}{64}+\frac{x+2}{63}+2=\frac{x+3}{62}+\frac{x+4}{61}+2\)
\(\left(\frac{x+1}{64}+\frac{64}{64}\right)+\left(\frac{x+2}{63}+\frac{63}{63}\right)=\left(\frac{x+3}{62}+\frac{62}{62}\right)+\left(\frac{x+4}{61}+\frac{61}{61}\right)\)
=> \(\frac{x+65}{64}+\frac{x+65}{63}=\frac{x+65}{62}+\frac{x+65}{61}\)\(\)
=> \(\frac{x+65}{64}+\frac{x+65}{63}-\frac{x+65}{62}-\frac{x+65}{61}=0\)
=> \(\left(x+65\right)\left(\frac{1}{64}+\frac{1}{63}-\frac{1}{62}-\frac{1}{61}\right)=0\)
Do \(\frac{1}{64}+\frac{1}{63}-\frac{1}{62}-\frac{1}{61}\ne0\)=> \(x+65=0\)
=> \(x=-65\)
b , Lm tương tự như Câu a
Chúc bn hok tốt
a) \(\frac{x+1}{65}+\frac{x+2}{64}=\frac{x+3}{63}+\frac{x+4}{62}\)
\(\Leftrightarrow\frac{x+1}{65}+\frac{x+2}{64}+2=\frac{x+3}{63}+\frac{x+4}{62}+2\)
\(\Leftrightarrow\left(\frac{x+1}{65}+1\right)+\left(\frac{x+2}{64}+1\right)=\left(\frac{x+3}{63}+1\right)+\left(\frac{x+4}{62}+1\right)\)
\(\Leftrightarrow\frac{x+1+65}{65}+\frac{x+2+64}{64}=\frac{x+3+63}{63}+\frac{x+4+62}{62}\)
\(\Leftrightarrow\frac{x+66}{65}+\frac{x+66}{64}=\frac{x+66}{63}+\frac{x+66}{62}\)
\(\Leftrightarrow\frac{x+66}{65}+\frac{x+66}{64}-\frac{x+66}{63}-\frac{x+66}{62}=0\)
\(\Leftrightarrow\left(x+66\right)\left(\frac{1}{65}+\frac{1}{64}-\frac{1}{63}-\frac{1}{62}\right)=0\)
\(\Leftrightarrow x+66=0\)
\(\Leftrightarrow x=-66\)
CÂu b) làm tương tự:
- Trừ 3 cho hai vế ( câu a) mk cộng 2 cho hai vế)
- Tách -3 = -1-1-1 rồi kết hợp với mỗi hạng tử
CỐ LÊN NHÉ
NẾU bạn KHÔNG HIỂU thì câu b) mik sẽ làm kĩ càng và rõ ràng hơn cho bạn hiểu
a,x-5/100+x-4/101+x-3/102=x-100/5+x-101/4+x-102/3
=>\(\dfrac{x-5}{100}-1+\dfrac{x-4}{101}-1+\dfrac{x-3}{102}-1=\dfrac{x-100}{5}-1+\dfrac{x-101}{4}-1+\dfrac{x-102}{3}-1\)
=>x-105=0
=>x=105