Tìm X biết: X:
X x 9 - X x 4= 12645
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X x ( 9 - 4 ) = 12645
X x 5 = 12645
X = 12645 : 5
X = 2529
Trả lời
\(x.\left(x+2\right).\left(x+4\right).\left(x+6\right)=9\)
\(\Leftrightarrow\left[x.\left(x+6\right)\right].\left[\left(x+2\right).\left(x+4\right)\right]=9\)
\(\Leftrightarrow\left(x^2+6x\right).\left(x^2+6x+8\right)=9\)
Đặt \(x^2+6x=t\) ta có
\(t.\left(t+8\right)=9\)
\(\Leftrightarrow t^2+8t-9=0\)
\(\Leftrightarrow t^2-t+9t-9=0\)
\(\Leftrightarrow t.\left(t-1\right)+9.\left(t-1\right)=0\)
\(\Leftrightarrow\left(t-1\right).\left(t+9\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}t-1=0\\t+9=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}t=1\\t=-9\end{cases}}\)
TH1 \(t=1\)
\(\Rightarrow x^2+6x=1\)
\(\Leftrightarrow x^2+6x-1=0\)
\(\Leftrightarrow x^2+6x+9-10=0\)
\(\Leftrightarrow\left(x+3\right)^2=10=\left(\pm\sqrt{10}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=\sqrt{10}\\x+3=-\sqrt{10}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-3+\sqrt{10}\\x=-3-\sqrt{10}\end{cases}}\)
TH2: \(t=-9\)
\(\Rightarrow x^2+6x=-9\)
\(\Leftrightarrow x^2+6x+9=0\)
\(\Leftrightarrow\left(x+3\right)^2=0\)
\(\Leftrightarrow x+3=0\)
\(\Leftrightarrow x=-3\)
Vậy \(x\in\left\{-3+\sqrt{10};-3-\sqrt{10};-3\right\}\)
\(x\left(x+2\right)\left(x+4\right)\left(x+6\right)=9\)
\(\Leftrightarrow x^4+12x^3+44x^2+48x=9\)
\(\Leftrightarrow x^4+12x^3+44x^2+48x-9=0\)
\(\Leftrightarrow\left(x^3+9x^2+17x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(x^2+6x-1\right)\left(x+3\right)^2=0\)
TH1 : Ta có : \(6^2-4.\left(-1\right)=36+4=40>0\)Suy ra : \(x_1=\frac{-6-\sqrt{40}}{2};x_2=\frac{-6+\sqrt{40}}{2}\)
TH2 : \(\left(x+3\right)^2=0\Leftrightarrow x+3=0\Leftrightarrow x=-3\)
khoan cach giua 2 so la :2-1=1
co tat ca so so hang hay x la :(10-1):1+1=10
(x+1)+(x+2)+.....................(x+10)=2015
x*10+[(1+10)*10/2]=2015
x*10+55 =2015
x*10 =2015-55
x*10 =1960
x =1960/10
x =196
Ta có:(x-3)(x2+3x+9)-x(x2-4)=1
=> x3-27-x3+4x=1
=>4x=28=>x=7
\(\left(x+9\right)+\left(x-2\right)+\left(x+7\right)+\left(x-4\right)+\left(x+5\right)+\left(x-6\right)+\left(x+3\right)+\left(x-8\right)+\left(x+1\right)=95\)
\(x+9+x-2+x+7+x-4+x+5+x-6+x+3+x-8+x+1=95\)
\(\left(x\times9\right)+\left(9-2+7-4+5-6+3-8+1\right)=95\)
\(x\times9+5=95\)
\(x\times9=95-5\)
\(x\times9=90\)
\(x=90:9\)
\(x=10\)
(x + 9) + (x - 2) + (x + 7) + (x - 4) + (x + 5) + (x - 6) + (x + 3) + (x - 8) + (x + 1) = 95
x + 9 + x - 2 + x + 7 + x - 4 + x + 5 + x - 6 + x + 3 + x - 8 + x + 1
(x + x + x + x + x + x + x + x + x) + (9 - 8 + 7 - 6 + 5 - 4 + 3 - 2 + 1) = 95
9 × x + 5 = 95
9 × x = 95 - 5
9 × x = 90
x = 90 : 9
x = 10
a: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(\Leftrightarrow x^3+8-x^3-2x=15\)
\(\Leftrightarrow2x=-7\)
hay \(x=-\dfrac{7}{2}\)
b: Ta có: \(\left(x-2\right)^3-\left(x-4\right)\left(x^2+4x+16\right)+6\left(x+1\right)^2=49\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+64+6\left(x+1\right)^2=49\)
\(\Leftrightarrow-6x^2+12x+56+6x^2+12x+6=49\)
\(\Leftrightarrow24x=-13\)
hay \(x=-\dfrac{13}{24}\)
\(x\cdot9-x\cdot4=12645\)
\(\Leftrightarrow x\left(9-4\right)=12645\)
\(\Rightarrow x\cdot5=12645\)
\(\Rightarrow x=12645:5\)
\(\Rightarrow x=2529\)