Tìm x, biết
5x + 12x = 13x
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\(5x+5x^2=43x^3\\ \Rightarrow43x^3-5x^2-5x=0\\ \Rightarrow x\left(43x^2-5x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\43x^2-5x-5=0\left(1\right)\end{matrix}\right.\\ \Delta\left(1\right)=25+4.5.43=885\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5+\sqrt{885}}{86}\\x=\dfrac{5-\sqrt{885}}{86}\end{matrix}\right.\)
\(A=x^4+6x^3+13x^2+12x+12\)
\(=\left(x^4+6x^3+19x^2+30x+25\right)-6x^2-18x-30+17\)
\(=\left(x^4+6x^3+19x^2+30x+25\right)-6\left(x^2+3x+5\right)+17\)
\(=\left(x^2+3x+5\right)^2-6\left(x^2+3x+5\right)+17\)
Đặt \(t=x^2+3x+5\)
Khi đó \(A=t^2-6t+17=t^2-2.t.3+9+8=\left(t-3\right)^2+8\ge8\)
Dấu "=" xảy ra <=> t - 3 = 0 <=> t = 3
<=> \(x^2+3x+5=3\Leftrightarrow x^2+3x+2=0\)
\(\Leftrightarrow x^2+x+2x+2=0\)
\(\Leftrightarrow x\left(x+1\right)+2\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-2\end{cases}}\)
Vậy AMin = 8 khi và chỉ khi x = -1 hoặc x = -2
1)Ta có:
aaabbb
= 111000 x a + 111 x b = 111 x (1000.a+b)
= a00b.111
=> đpcm
a)\(5x^2=13x\Leftrightarrow5x^2-13x=0\Leftrightarrow x\left(5x-13\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\5x-13=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=\frac{13}{5}\end{array}\right.\)
b)\(6x^4=9x^3\Leftrightarrow6x^4-9x^3=0\Leftrightarrow3x^3\left(2x-3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}3x^3=0\\2x-3=0\end{array}\right.\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=\frac{3}{2}\end{array}\right.\)
c)\(\left(x-2\right)^2-4x^2-12x-9=0\)
\(\Leftrightarrow\left(x-2\right)^2=4x^2+12x+9\)
\(\Leftrightarrow\left(x-2\right)^2=\left(2x+3\right)^2\)
\(\Leftrightarrow x-2=2x+3\)
\(\Leftrightarrow-x=5\Leftrightarrow x=-5\)
a)
\(\frac{x\left(x+1\right)}{2}=36\)(quy tắc tính tổng)
=>\(x\left(x+1\right)=72\)
=>\(x\left(x+1\right)=8.9\)
=>x=8
b)\(12x+13x=2000\Rightarrow\left(12+13\right)x=2000\Rightarrow25x=2000\Rightarrow x=80\)
c)
(x-1)(x-5)=0
\(\Rightarrow\orbr{\begin{cases}x-1=0\\x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=5\end{cases}}}\)
cái ngoặc vuông là "hoặc " nhé
Ta có: \(x^2=y^2+z^2\)
\(\Leftrightarrow x^2-y^2=z^2\)
\(\Leftrightarrow25\left(x-y\right)\left(x+y\right)=25z^2\)
\(\Leftrightarrow\left(25x-25y\right)\left(x+y\right)=25z^2\)
\(\Leftrightarrow\left(13x-12y+12x-13y\right)\left(13x-12y-12x+13y\right)=25z^2\)
\(\Leftrightarrow\left(13x-12y\right)^2-\left(12x-13y\right)^2=25z^2\)
\(\Leftrightarrow\left(13x-12y\right)^2-\left(5z\right)^2=\left(12x-13y\right)^2\)
\(\Leftrightarrow\left(13x-12y-5z\right)\left(13x-12y+5z\right)=\left(12x-13y\right)^2\)(ĐPCM).
Chia cả 2 vế cho 13x ta được
\(\left(\frac{5}{13}\right)^x+\left(\frac{12}{13}\right)^x=1\)
Xét x=2 thì thỏa mãn
Xét \(0\le x< 2\)thì không có số thỏa mãn
Xét x<0 thì VT>1 (vô lí)
Xét x>2 thì \(VT< \frac{5^2+12^2}{13^2}=1\)(vô lí)
Vậy x=2