\(x-6:2-\left(48-24\right):12-13=0\)
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b) \(\left[x.\left(x-1\right)\right].\left[\left(x-2\right).\left(x+1\right)\right]=24\)
\(\Leftrightarrow\left(x^2-x\right).\left(x^2-x-2\right)=24\)
Đặt \(x^2-x=k\), ta có:
\(k.\left(k-2\right)=24\)
\(\Leftrightarrow k^2-2k+1=25\)
\(\Leftrightarrow\left(k-1\right)^2=5^2\Leftrightarrow\orbr{\begin{cases}k-1=5\\k-1=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}k=6\\k=-4\end{cases}}}\)
\(k=6\Rightarrow x^2-x=6\Rightarrow x^2-x-6=0\)
\(\Rightarrow x^2-3x+2x-6=0\Rightarrow x.\left(x-3\right)+2.\left(x-3\right)=0\)
\(\Rightarrow\left(x+2\right).\left(x-3\right)=0\Rightarrow\orbr{\begin{cases}x=-2\\x=3\end{cases}}\)
\(k=-4\Rightarrow x^2-x+4=0\Rightarrow x^2-x+\frac{1}{4}+\frac{15}{4}=0\Rightarrow\left(x-\frac{1}{2}\right)^2=-\frac{15}{4}\left(\text{loại}\right)\)
c)\(x^4+2x^3+5x^2+4x-12=0\)
\(\Leftrightarrow x^4+2x^3+2x^2+4x+3x^2-12=0\)
\(\Leftrightarrow x^3.\left(x+2\right)+2x.\left(x+2\right)+3.\left(x^2-2^2\right)=0\)
\(\Leftrightarrow\left(x+2\right).\left(x^3+5x-6\right)=0\)
\(\Leftrightarrow\left(x+2\right).\left(x^3-x^2+x^2-x+6x-6\right)=0\)
\(\Leftrightarrow\left(x+2\right).\left[x^2.\left(x-1\right)+x.\left(x-1\right)+6.\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x+2\right).\left(x-1\right).\left(x^2+x+6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=1\end{cases}\text{vì }x^2+x+6>0\left(\text{tự c/m}\right)}\)
p/s: bn tự kết luận nha :))
\(8,1-\left(x-6\right)=4\left(2-2x\right)\)
\(\Leftrightarrow1-x+6=8-8x\)
\(\Leftrightarrow-x+8x=8-1-6\)
\(\Leftrightarrow7x=1\)
\(\Leftrightarrow x=\dfrac{1}{7}\)
\(9,\left(3x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)
\(10,\left(x+3\right)\left(x^2+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\varnothing\end{matrix}\right.\)
`8)1-(x-5)=4(2-2x)`
`<=>1-x+5=8-6x`
`<=>5x=2<=>x=2/5`
`9)(3x-2)(x+5)=0`
`<=>[(x=2/3),(x=-5):}`
`10)(x+3)(x^2+2)=0`
Mà `x^2+2 > 0 AA x`
`=>x+3=0`
`<=>x=-3`
`11)(5x-1)(x^2-9)=0`
`<=>(5x-1)(x-3)(x+3)=0`
`<=>[(x=1/5),(x=3),(x=-3):}`
`12)x(x-3)+3(x-3)=0`
`<=>(x-3)(x+3)=0`
`<=>[(x=3),(x=-3):}`
`13)x(x-5)-4x+20=0`
`<=>x(x-5)-4(x-5)=0`
`<=>(x-5)(x-4)=0`
`<=>[(x=5),(x=4):}`
`14)x^2+4x-5=0`
`<=>x^2+5x-x-5=0`
`<=>(x+5)(x-1)=0`
`<=>[(x=-5),(x=1):}`
Bài 2:
a)\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9x-18}+6\sqrt{\dfrac{x-2}{81}}=-4\) (đk: \(x\ge2\))
\(\Leftrightarrow\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9\left(x-2\right)}+\dfrac{6}{\sqrt{81}}\sqrt{x-2}=-4\)
\(\Leftrightarrow\dfrac{1}{3}\sqrt{x-2}-2\sqrt{x-2}+\dfrac{2}{3}\sqrt{x-2}=-4\)
\(\Leftrightarrow-\sqrt{x-2}=-4\) \(\Leftrightarrow x-2=16\)
\(\Leftrightarrow x=18\) (thỏa)
Vậy...
b)\(\sqrt{9x^2+12x+4}=4x\)(Đk:\(9x^2+12x+4\ge0\))
\(\Leftrightarrow\left\{{}\begin{matrix}4x\ge0\\9x^2+12x+4=16x^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\-7x^2+12x+4=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\-7x^2+14x-2x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left(x-2\right)\left(-7x-2\right)=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left[{}\begin{matrix}x=2\\x=-\dfrac{2}{7}\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow x=2\) (tm đk)
Vậy...
c) \(\sqrt{x-2\sqrt{x-1}}=\sqrt{x-1}\) (đk: \(x\ge1\))
\(\Leftrightarrow x-2\sqrt{x-1}=x-1\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{1}{2}\) \(\Leftrightarrow x=\dfrac{5}{4}\) (tm)
Vậy...
\(A=\dfrac{75x\left(12+x\right)}{\left(12+4x\right)^2}\);\(A>0\forall x>0\)
Gọi \(A_0\in MGT\) của A
\(\Rightarrow A_0=\dfrac{75x\left(12+x\right)}{\left(12+4x\right)^2}\) có nghiệm
\(\Rightarrow A_0\left(12+4x\right)^2=75x\left(12+x\right)\)
\(\Leftrightarrow x^2\left(16A_0-75\right)+x\left(96A_0-900\right)+144A_0=0\) có nghiệm
\(\Leftrightarrow\Delta\ge0\Leftrightarrow-4A_0+25\ge0\)\(\Leftrightarrow A_0\le\dfrac{25}{4}\)
\(\Rightarrow maxA=\dfrac{25}{4}\)
\(\Leftrightarrow\)\(x-\left(\frac{13x}{18}-\frac{4}{18}\right)=\frac{4}{9}\)
\(\Leftrightarrow\)\(\frac{18x}{18}-\frac{13x}{18}+\frac{4}{18}=\frac{4}{9}\)
\(\Leftrightarrow\)\(\frac{5x}{18}=\frac{4}{9}-\frac{4}{18}\)
\(\Leftrightarrow\)\(\frac{5x}{18}=\frac{2}{9}\)
\(\Leftrightarrow\)\(5x=\frac{18.2}{9}\)
\(\Leftrightarrow\)\(5x=4\)
\(\Leftrightarrow\)\(x=\frac{4}{5}\)
x - 6:2 - (48 - 24) : 12 - 13 = 0
x - 3 - 24 : 12 - 13 = 0
x - 3 - 12 - 13 = 0
x - 3 - 12 = 0 + 13
x - 3 - 12 = 13
x - 3 = 13 + 12
x - 3 = 25
x = 25 + 3
x = 28
đề bài sai