tim x 1+2+3+.......+.x= 45
tim x 1+2+3+...+x=45
Ta có (x+1)*x/2=45
=> (x+1)*x=90=9*10
=> x=9
Ta có: 1 + 2 + 3 + ... + x = 45
\(\Rightarrow\) \(\dfrac{\left[\left(x-1\right)+1\right]}{2}\). ( x + 1 ) = 45
Lưu ý: cái này là dựa vào công thức tính tổng các số hạng nha:
Công thức: muốn tính tổng các số hạng:
[(số cuối - số đầu): khoảng cách giữa hai số + 1 ] : 2 . (số cuối + số đầu )
\(\Rightarrow\) x : 2 .( x + 1 ) = 45
\(\Rightarrow\) x . ( x + 1 ) = 45 . 2
\(\Rightarrow\) x . ( x + 1 ) = 90 = 9 . 10
\(\Rightarrow\) x = 9
Chúc bn học tốt !!!
tim x 1+2+3+...+x=45
Ta có :
1 + 2 + 3 +...+ x = 45
=> x . ( x + 1 ) : 2 = 45
=> x . ( x + 1 ) = 45 . 2 = 90
Ta có :
x . ( x+ 1 ) = 90 = 9 x 10
=> x = 9
Vậy x = 9
Bạn dùng công thức tính tổng dãy số có quy luật là ra ak
Ta có \(1+2+3+....+x=45\)
\(\Rightarrow\frac{\left(x+1\right).x}{2}=45\)
\(\Rightarrow x.\left(x+1\right)=90\)
\(\Rightarrow x.\left(x+1\right)=9.10\)
\(\Rightarrow x=9\)
Vậy x = 9
tim x
2 . 3x-1 = 45
2 x 3x-1 = 45
3x-1 = 45 : 2 =24,5
=>Không tồn tại x.
2 . 3x - 1 = 45
3x - 1 = 45 : 2
x = không tồn tại
Câu 2: (2 điểm) Tim x, biết:
1) (x−1)/2009+(x−2)/2008=(x−3)/2007+(x−4)/2006
2) (59−x)/41+(57−x)/43+(55−x)/45+(53−x)/47+(51−x)/49=−5
1, \(\dfrac{x-1}{2009}+\dfrac{x-2}{2008}=\dfrac{x-3}{2007}+\dfrac{x-4}{2006}\)
\(\Leftrightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\) ( Trừ mỗi vế cho 2 ta được phương trình như này nhé ! )
\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}=\dfrac{x-2010}{2007}+\dfrac{x-2010}{2006}\)
\(\Leftrightarrow\dfrac{x-2010}{2009}+\dfrac{x-2010}{2008}-\dfrac{x-2010}{2007}-\dfrac{x-2010}{2006}=0\)
\(\Leftrightarrow\left(x-2010\right)\left(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)=0\)
Do \(\dfrac{1}{2009}+\dfrac{1}{2008}-\dfrac{1}{2007}-\dfrac{1}{2006}\ne0\) nên \(x-2010=0\Leftrightarrow x=2010\)
2, \(\dfrac{59-x}{41}+\dfrac{57-x}{43}+\dfrac{55-x}{45}+\dfrac{53-x}{47}+\dfrac{51-x}{49}=-5\)
\(\left(\dfrac{59-x}{41}+1\right)+\left(\dfrac{57-x}{43}+1\right)+\left(\dfrac{55-x}{45}+1\right)+\left(\dfrac{53-x}{47}+1\right)+\left(\dfrac{51-x}{49}+1\right)=0\)
\(\Leftrightarrow\dfrac{100-x}{41}+\dfrac{100-x}{43}+\dfrac{100-x}{45}+\dfrac{100-x}{47}+\dfrac{100-x}{49}=0\) \(\Leftrightarrow\left(100-x\right)\left(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\right)=0\) Do \(\dfrac{1}{41}+\dfrac{1}{43}+\dfrac{1}{45}+\dfrac{1}{47}+\dfrac{1}{49}\ne0\) nên \(100-x=0\Leftrightarrow x=100\)
Tim x: 6^2-(x+3)=45
125-5.(3x-1)=5^5.5^3
4^x+1+4^0=65
70-5.(x-3)=5.3^2
\(6^2-\left(x+3\right)=45\)
\(36-\left(x+3\right)=45\)
\(x+3=35-45\)
\(x+3=-10\)
\(x=-13\)
\(6^2-\left(x+3\right)=45\)
\(\Rightarrow36-\left(x+3\right)=45\)
\(\Rightarrow x+3=36-45\)
\(\Rightarrow x+3=-9\)
\(\Rightarrow x=-9-3=-12\)
\(125-5\left(3x-1\right)=5^5.5^3\)
\(\Rightarrow5^3-5\left(3x-1\right)=5^5.5^3\)
\(\Rightarrow5^3-15x+5=5^5.5^3\)
\(\Rightarrow5\left(5^2-3x+1\right)=5^5.5^3\)
\(\Rightarrow5^2-3x+1=5^5.5^3:5\)
\(\Rightarrow25-3x+1=5^7\)
Từ đây làm nốt nhé
\(4^x+1+4^0=65\)
\(\Rightarrow4^{x+1}+1=65\)
\(\Rightarrow4^{x+1}=64\)
\(\Rightarrow4^{x+1}=4^3\)
\(\Rightarrow x+1=3\)
\(\Rightarrow x=2\)
\(70-5\left(x-3\right)=5.3^2\)
\(\Rightarrow70-5\left(x-3\right)=5.9=45\)
\(\Rightarrow5\left(x-3\right)=70-45=25\)
\(\Rightarrow x-3=25:5=5\)
\(\Rightarrow x=8\)
Chúc em học tốt hơn nhé!!
1, tim x
a, (3.x+4):12+17=19
b, 45: (2.x+1) -3=0
c, 36+72.(x-1)=396
d,470-(2.x+3).11=217
a. \(\Leftrightarrow\) (3.x+4):12 = 19-17
\(\Leftrightarrow\)(3.x+4):12 = 2
\(\Leftrightarrow\)3.x+4 = 24 \(\Leftrightarrow\)3.x = 20 \(\Leftrightarrow\)x = \(\frac{20}{3}\)
b. \(\Leftrightarrow\)45: (2.x+1) = 3 \(\Leftrightarrow\)2.x+1 = 15 \(\Leftrightarrow\)2.x = 14 \(\Leftrightarrow\)x = 7
c. \(\Leftrightarrow\)72.(x+1) = 360 \(\Leftrightarrow\)x+1 = 5\(\Leftrightarrow\)x = 4
d.\(\Leftrightarrow\)(2.x+3).11 = 253\(\Leftrightarrow\)2.x+3 = 23\(\Leftrightarrow\)2.x = 20\(\Leftrightarrow\)x = 10
3022x-6+45=46
Tim so tu nhien x biet
(x+1)+(x+2)+(x+3)+....+(x+100)=5950
a) Ta có: 302^(2x-6) +45 = 46
=> 302^(2x-6) = 1 = 302^0
=> 2x-6 = 0
=> 2x = 6
=> x = 3
b) Ta có: (x+1)+(x+2)+(x+3)+....+(x+100)=5950
=> 100x+ (1+2+3+4+...+100) = 5950
=> 100x + 5050 = 5950
=> 100x = 900
=> x = 9
Nhấn đúng cho mk nha!!!!!!!!
Tim x, biet:
\(\frac{x+1}{49}+\frac{x+2}{48}+\frac{x+3}{47}+\frac{x+4}{46}+\frac{x+5}{45}=-5\)
Ta có :
\(\frac{x+1}{49}+\frac{x+2}{48}+\frac{x+3}{47}+\frac{x+4}{46}+\frac{x+5}{45}=-5\)
\(\Leftrightarrow\)\(\left(\frac{x+1}{49}+1\right)+\left(\frac{x+2}{48}+1\right)+\left(\frac{x+3}{47}+1\right)+\left(\frac{x+4}{46}+1\right)+\left(\frac{x+5}{45}+1\right)=-5+5\)
\(\Leftrightarrow\)\(\frac{x+50}{49}+\frac{x+50}{48}+\frac{x+50}{47}+\frac{x+50}{46}+\frac{x+50}{45}=0\)
\(\Leftrightarrow\)\(\left(x+50\right)\left(\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{46}+\frac{1}{45}\right)=0\)
Vì \(\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{46}+\frac{1}{45}\ne0\)
Nên \(x+50=0\)
\(\Rightarrow\)\(x=-50\)
Vậy \(x=-50\)
Chúc bạn học tốt ~
5/x = x/ 45
x+1 /12 = 3/ x+1
tim x biet
\(\frac{5}{x}=\frac{x}{45}\)
=> \(x^2=225\)
<=> \(x=15\)hoặc \(x=-15\)
\(x+\frac{1}{12}=\frac{3}{x+1}\)
<=> \(\frac{12x+1}{12}=\frac{3}{x+1}\)
=> \(12x^2+12x+x+1=36\)
<=> \(12x^2+13x+1=36\)
<=> \(12x^2+13x-35=0\)
<=> \(\left(x-\frac{5}{4}\right)\left(x+\frac{7}{3}\right)=0\)
<=> \(\hept{\begin{cases}x-\frac{5}{4}=0\\x+\frac{7}{3}=0\end{cases}}\)
<=>\(\hept{\begin{cases}x=\frac{5}{4}\\x=\frac{-7}{3}\end{cases}}\)
a)\(\frac{5}{x}=\frac{x}{45}\)
\(\Rightarrow x^2=5.45\)
\(\Rightarrow x^2=225=\orbr{\begin{cases}15^2\\\left(-15\right)^2\end{cases}}\)
\(\Rightarrow x=\orbr{\begin{cases}15\\-15\end{cases}}\)
b) \(\frac{x+1}{12}=\frac{3}{x+1}\)
\(\Rightarrow\left(x+1\right)^2=36\)
\(\Rightarrow x+1=\orbr{\begin{cases}6\\-6\end{cases}}\)
+) \(\frac{5}{x}=\frac{x}{45}\)
\(\Rightarrow x\times x=5\times45\)
\(\Rightarrow x^2=225\)
\(\Rightarrow\orbr{\begin{cases}x^2=15^2\\x^2=\left(-15\right)^2\end{cases}\Rightarrow}\orbr{\begin{cases}x=15\\x=-15\end{cases}}\)
Vậy x = 15 hoặc x = -15
+) \(\frac{x+1}{12}=\frac{3}{x+1}\)
\(\Rightarrow\left(x+1\right)\times\left(x+1\right)=3\times12\)
\(\Rightarrow\left(x+1\right)^2=36\)
\(\Rightarrow\orbr{\begin{cases}\left(x+1\right)^2=6^2\\\left(x+1\right)^2=\left(-6\right)^2\end{cases}\Rightarrow}\orbr{\begin{cases}x+1=6\\x+1=-6\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=-7\end{cases}}\)
Vậy x = 5 hoặc x = -7
_Chúc bạn học tốt_