tìm x, biết:
x+x+x+x+x+x+x+x+x+x+x+x+x+x+x+x+x+x+x+x+x+x+x+x+x=380125
45.2+2+x+45.2=2525255
Tìm X, biết:
X + (X + 1) + (X + 2) + (X + 3) + ...+ (X + 19) = 950
=>20x+190=950
=>20x=760
hay x=38
`20x+190=950`
`20x=760`
`x= 760: 20`
`x= 38`
20 x X = 950 - (1 + 2 + 3 + 4 + 5 + ... + 19)
20 x X = 950 - 190
20 x X = 760
X = 760 : 20
X = 38
Tìm x biết:
x(x+5)(x-5)-(x+2)(x^2-2x+4)=42
\(x\left(x+5\right)\left(x-5\right)-\left(x+2\right)\left(x^2-2x+4\right)=42\)
\(\Leftrightarrow x\left(x^2-25\right)-\left(x^3+8\right)=42\)
\(\Leftrightarrow x^3-25x-x^3-8=42\)
\(\Leftrightarrow-25x-8=42\)
\(\Leftrightarrow-25x=42+8\)
\(\Leftrightarrow-25x=50\)
\(\Leftrightarrow x=-\dfrac{50}{25}=-2\)
Tìm x biết:X x 7 + X x 9 - X - X x 5 = 790
Tìm x > 0 biết:x(x^3y-x)-x^2(x^2y-2)=4
Tìm x > 0 biết:x(x^3y-x)-x^2(x^2y-2)=4
Tìm x,biết:
x(x+1)-(x-1)(x+2)=8
x(x+1)-(x-1)(x+2)=8
\(\Leftrightarrow\)\(x^2+x-x^2-2x+x+2=8\)
\(\Leftrightarrow0x=6\left(ptvn\right)\)
\(\Rightarrow S=\varnothing\)
Tìm x, biết:
x+(x+1)+(x+2)+...+(x+30)=1240
`x+(x+1)+(x+2)+...+(x+30)=1240`
`=> (x + x + x + ... + x) + (1 + 2 + 3 +... + 30) = 1240`
`=> 31x + 465 = 1240`
`=> 31 x = 1240 - 465`
`⇒ 31x = 775`
`⇒ x = 775 : 31`
`⇒ x = 25`
\(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+30\right)=1240\\ \left(x+x+...+x\right)+\left(1+2+...+30\right)=1240\\ 31x+465=1240\\ 31x=1240-465\\ 31x=775\\ x=775:31\\ x=25\)
\(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+30\right)=1240\)
\(\Leftrightarrow\left(x+x+...+x\right)+\left(1+2+...+30\right)=1240\)
\(\Leftrightarrow31x+465=1240\)
\(\Leftrightarrow x=25\)
Tìm xϵZ biết:
x-{[-x+(x+3)]} - [(x+3)-(x-2)]=0
\(\Rightarrow x-\left(-x+x+3\right)-\left(x+3-x+2\right)=0\\ \Rightarrow x-3-5=0\\ \Rightarrow x=8\)
Sai rồi bạn ơi, bằng -2 mới đúng,bn thử lại coi
tìm x biết:x-1/x+2=x-2/x+3
x³ - x² - x = 1/3
<=> x³ = x² + x + 1/3
<=> 3x³ = 3(x² + x + 1/3)
<=> 3x³ = 3x² + 3x + 1
<=> 3x³ + x³ = x³ + 3x² + 3x + 1
<=> 4x³ = (x + 1)³
<=> ³√(4x³) = ³√(x + 1)³
<=> ³√4.x = x + 1
<=> ³√4.x - x = 1
<=> x(³√4 - 1) = 1
<=> x = 1/(³√4 - 1)
Ta có \(\frac{x-1}{x+2}=\frac{x-2}{x+3}\)
\(\Rightarrow\left(x-1\right)\left(x+3\right)=\left(x+2\right)\left(x-2\right)\)
\(\Rightarrow x^2+2x-3=x^2-4\)
\(\Rightarrow x^2-x^2+2x=-4+3\)
\(\Rightarrow2x=-1\)
\(\Rightarrow x=-\frac{1}{2}\)
Vậy \(x=-\frac{1}{2}\)