Tìm x biết
1+2+3+4+.....+x=45
Tìm x biết (x+1)+(x+2)+(x+3)+(x+4)+(x+5)=45
x+1 + x+2 + x+3 + x +4 + x+5 = 45
x + x + x + x + x = 45 -1 -2 -3 -4 -5
5x = 30
x =6
tìm x biết:(1/1*2*3+1/2*3*4+...+1/8*9*10)*x=22/45
tìm x biết:
1) 14- (5-8+3-x) = |-7+10|
2) -19 - ?(3+x-5) = |4-15+2|
3) 45- (x+45-3) = |-7+3|
Bài 1 : Tìm x biết :
x + 4/5.9 + 4/ 9.13 + 4/ 13 .15 + ... + 4/41.45 = -37/45
Bài 2 : x - 20/11.13 - 20/13.15 - ...-20/53.55= 3/11
Bài 3 : Tìm x biết
1/21 + 1/28 + 1/36 + ...+ 2/ x(x+1)=2/9
tìm số tự nhiên x, biết:(1/1*2*3 1/2*3*4 ..... 1/8*9*10)*x=23/45
tìm y biết 2/3 x 4/y x 1/5=4/45
3/4 x y/5=5/2 x 9/10
\(\frac{2}{3}x\frac{4}{y}=\frac{4}{45}:\frac{1}{5}\)\(=\frac{4}{15}\)
\(\frac{4}{y}=\frac{4}{15}:\frac{2}{3}\)\(=\frac{2}{5}\)
y=4:\(\frac{2}{5}\) y=10
\(\frac{3}{4}x\frac{y}{5}=\frac{15}{4}\)
\(\frac{y}{5}=\frac{15}{4}:\frac{3}{4}=5\)
y=5x5=25
Bài 2:Tìm x,biết:
1, 4/5-x=1/3
2, 2/3+5/3x=5/7
3, -12/13x+5=5 1/13
4, x:(-2 1/4)+1,5=-3/4
5,GTTĐ của x -3/4 - 1/4=0
6, 6-GTTĐ của 1/2-x = 2/3
7, (x-1)2=4/9
1) x = 4/5 - 1/3
x = 7/15
2) 5/3.x=1/21
x=1/35
3) -12/13.x = 1/13
x=-1/12
7) th1: x-1=2/3
x = 5/3
Th2: x - 1 = -2/3
x=1/3
4) tìm x, biết
A) 1+2+...+x=45
B) 1+3+5+...+x=36
a) \(1+2+...+x=45\)
\(\Rightarrow\frac{x\left(x+1\right)}{2}=45\)
\(\Rightarrow x\left(x+1\right)=90\)
\(\Rightarrow x\left(x+1\right)=9\cdot10\)
VẬy x = 9
Câu 2: Tìm x biết:
a. \(\sqrt{x-1}=2\)
b. \(\sqrt{3x+1}=\sqrt{4x-3}\)
c. \(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=6\)
d. \(\sqrt{x^2-4x+4}=\sqrt{6+2\sqrt{5}}\)
\(a,\Leftrightarrow x-1=4\Leftrightarrow x=5\\ b,\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\3x+1=4x-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{3}{4}\\x=4\left(tm\right)\end{matrix}\right.\Leftrightarrow x=4\\ c,ĐK:x\ge-5\\ PT\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\\ \Leftrightarrow3\sqrt{x+5}=6\\ \Leftrightarrow\sqrt{x+5}=3\\ \Leftrightarrow x+5=9\\ \Leftrightarrow x=4\left(tm\right)\)
\(d,\Leftrightarrow\sqrt{\left(x-2\right)^2}=\sqrt{\left(\sqrt{5}+1\right)^2}\\ \Leftrightarrow\left|x-2\right|=\sqrt{5}+1\\ \Leftrightarrow\left[{}\begin{matrix}x-2=\sqrt{5}+1\\2-x=\sqrt{5}+1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{5}+3\\x=1-\sqrt{5}\end{matrix}\right.\)