Tim x biet -1+3+(-5)+7+...+x=600
giup minh voi
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\(\frac{5}{3}-\frac{1}{3}:\left(1-x\cdot\frac{1}{3}\right)=\frac{7}{6}\)
=> \(\frac{1}{3}:\left(1-x\cdot\frac{1}{3}\right)=\frac{5}{3}-\frac{7}{6}\)
=> \(\frac{1}{3}:\left(1-x\cdot\frac{1}{3}\right)=\frac{1}{2}\)
=> \(\left(1-x\cdot\frac{1}{3}\right)=\frac{1}{3}:\frac{1}{2}=\frac{1}{3}\cdot2=\frac{2}{3}\)
=> \(1-\frac{x}{3}=\frac{2}{3}\)
=> \(\frac{x}{3}=1-\frac{2}{3}=\frac{1}{3}\)
=> x = 1
\(3-\left(x:\frac{1}{2}+\frac{3}{2}\right)-\frac{1}{4}=\frac{1}{2}\)
=> \(3-\left(x:\frac{1}{2}+\frac{3}{2}\right)=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}\)
=> \(x:\frac{1}{2}+\frac{3}{2}=3-\frac{3}{4}=\frac{9}{4}\)
=> \(x:\frac{1}{2}=\frac{9}{4}-\frac{3}{2}\)
=> \(x:\frac{1}{2}=\frac{3}{4}\)
=> \(x=\frac{3}{4}\cdot\frac{1}{2}=\frac{3}{8}\)
\(\frac{5}{3}-\frac{1}{3}:\left(1-x\times\frac{1}{3}\right)=\frac{7}{6}\)
\(\frac{1}{3}:\left(1-x\times\frac{1}{3}\right)=\frac{5}{3}-\frac{7}{6}\)
\(\frac{1}{3}:\left(1-x\times\frac{1}{3}\right)=\frac{1}{2}\)
\(1-x\times\frac{1}{3}=\frac{1}{3}:\frac{1}{2}\)
\(1-x\times\frac{1}{3}=\frac{2}{3}\)
\(x\times\frac{1}{3}=1-\frac{2}{3}\)
\(x\times\frac{1}{3}=\frac{1}{3}\)
\(x=\frac{1}{3}:\frac{1}{3}\)
\(x=1\)
\(3-\left(x:\frac{1}{2}+\frac{3}{2}\right)-\frac{1}{4}=\frac{1}{2}\)
\(3-\left(x:\frac{1}{2}+\frac{3}{2}\right)=\frac{1}{2}+\frac{1}{4}\)
\(3-\left(x:\frac{1}{2}+\frac{3}{2}\right)=\frac{3}{4}\)
\(x:\frac{1}{2}+\frac{3}{2}=3-\frac{3}{4}\)
\(x:\frac{1}{2}+\frac{3}{2}=\frac{9}{4}\)
\(x:\frac{1}{2}=\frac{9}{4}-\frac{3}{2}\)
\(x:\frac{1}{2}=\frac{3}{4}\)
\(x=\frac{3}{4}\times\frac{1}{2}\)
\(x=\frac{3}{8}\)
\(\left(x-3\right)\left(x-5\right)+1=0\)
\(\Leftrightarrow x^2-5x-3x+15+1=0\)
\(\Leftrightarrow x^2-8x+16=0\)
\(\Leftrightarrow\left(x-4\right)^2=0\)
\(\Leftrightarrow x-4=0\)
\(\Leftrightarrow x=4\)
Vậy \(x=4\)
\(\left(x-3\right)\left(x-5\right)+1=0\)
\(\Rightarrow x^2-3x-5x+15+1=0\)
\(\Rightarrow x^2-8x+16=0\)
\(\Rightarrow x^2-2x.4+4^2=0\)
\(\Rightarrow\left(x-4\right)^2=0\)
\(\Rightarrow x-4=0\)
\(\Rightarrow x=4\)
Vậy \(x=4\)
\(∘backwin\)
\(a ) ( x + 1 ) + ( x + 2 ) + ( x + 3 ) + ... + ( x + 100 ) = 5750\)
\( ( x + x + x + ... + x ) + ( 1 + 2 + 3 + ... + 100 ) = 5750 \)
\( 100 x + ( 1 + 100 ) ×100 : 2 = 5750\)
\(100 x + 5050 = 5750\)
\( 100 x = 5750 − 5050\)
\(100 x = 700\)
\(x = 700 : 100\)
\(x = 7\)
\(b,\) \(B=\)\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2021^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2020}+2021\)
\( B < 1 -\)\(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2020}-\dfrac{1}{2021}\)
\(B<1-\)\(\dfrac{1}{2021}\)
\(B<\)\(\dfrac{2020}{2021}\)
\(\dfrac{2020}{2021}< 1\)
\(B<1\)
a) (x+1) +(x+2 ) + ...+(x+100)=5750
= 100x + (1+2+3+...+100) = 5750
=100x + 5050 = 5750
--> 100x = 5750-5050=700
--> x=7
a)(x - 2)(y + 3)= 5
Vì x,y là các số nguyên => x-2,y+3 là các ước nguyên của 5
Ta có bảng sau:
x-2 | 1 | 5 | -1 | -5 |
y+3 | 5 | 1 | -5 | -1 |
x | 3 | 7 | 1 | -3 |
y | 2 | -2 | -8 | -4 |
b) (x + 1)(y - 5) =-7
Vì x,y là các số nguyên => x+1,y-5 là các ước nguyên của -7
Ta có bảng sau:
x+1 | 1 | -7 | -1 | 7 |
y-5 | -7 | 1 | 7 | -1 |
x | 0 | -8 | -2 | 6 |
y | -2 | 6 | 12 | 4 |
Chúc bạn học tốt!