18-12x=5x-(3-4)
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a: \(\Leftrightarrow\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\cdot3\sqrt{x-2}+6\cdot\dfrac{\sqrt{x-2}}{9}=-4\)
\(\Leftrightarrow\sqrt{x-2}=4\)
=>x-2=16
hay x=18
b: \(\Leftrightarrow\left|3x+2\right|=4x\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+2=4x\left(x>=-\dfrac{2}{3}\right)\\3x+2=-4x\left(x< -\dfrac{2}{3}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=-\dfrac{2}{7}\left(nhận\right)\end{matrix}\right.\)
c: \(\Leftrightarrow3\sqrt{x-2}-2\sqrt{x-2}+3\sqrt{x-2}=40\)
\(\Leftrightarrow4\sqrt{x-2}=40\)
=>x-2=100
hay x=102
d: =>5x-6=9
hay x=3
\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9x-18}+6\sqrt{\dfrac{x-2}{81}}=-4\) (đk: x≥2)
\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9\left(x-2\right)}+6\sqrt{\dfrac{1}{81}\left(x-2\right)}=-4\)
\(\dfrac{1}{3}\sqrt{x-2}-2\sqrt{x-2}+\dfrac{2}{3}\sqrt{x-2}=-4\)
\(\dfrac{1}{3}\sqrt{x-2}-\dfrac{4}{3}\sqrt{x-2}=-4\)
\(-\sqrt{x-2}=-4\)
\(\sqrt{x-2}=4\)
\(\left|x-2\right|=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=16\\x-2=-16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=18\left(TM\right)\\x=-14\left(L\right)\end{matrix}\right.\)
a: ĐKXĐ: \(x\in R\)
\(\sqrt{\left(2x+3\right)^2}=5\)
=>|2x+3|=5
=>\(\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)
b: ĐKXĐ: \(x\in R\)
\(\sqrt{9\left(x-2\right)^2}=18\)
=>\(\sqrt{9}\cdot\sqrt{\left(x-2\right)^2}=18\)
=>\(3\cdot\left|x-2\right|=18\)
=>\(\left|x-2\right|=6\)
=>\(\left[{}\begin{matrix}x-2=6\\x-2=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)
c: ĐKXĐ: x>=2
\(\sqrt{9x-18}-\sqrt{4x-8}+3\sqrt{x-2}=40\)
=>\(3\sqrt{x-2}-2\sqrt{x-2}+3\sqrt{x-2}=40\)
=>\(4\sqrt{x-2}=40\)
=>\(\sqrt{x-2}=10\)
=>x-2=100
=>x=102(nhận)
d: ĐKXĐ: \(x\in R\)
\(\sqrt{4\left(x-3\right)^2}=8\)
=>\(\sqrt{\left(2x-6\right)^2}=8\)
=>|2x-6|=8
=>\(\left[{}\begin{matrix}2x-6=8\\2x-6=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=14\\2x=-2\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=7\left(nhận\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)
e: ĐKXĐ: \(x\in R\)
\(\sqrt{4x^2+12x+9}=5\)
=>\(\sqrt{\left(2x\right)^2+2\cdot2x\cdot3+3^2}=5\)
=>\(\sqrt{\left(2x+3\right)^2}=5\)
=>|2x+3|=5
=>\(\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)
f: ĐKXĐ:x>=6/5
\(\sqrt{5x-6}-3=0\)
=>\(\sqrt{5x-6}=3\)
=>\(5x-6=3^2=9\)
=>5x=6+9=15
=>x=15/5=3(nhận)
d, \(x^8+x^7+1\)
\(=x^8-x^2+x^7-x+x^2+x+1\)
\(=x^2\left(x^6-1\right)+x\left(x^6-1\right)+x^2+x+1\)
\(=x^2\left(x^3-1\right)\left(x^3+1\right)+x\left(x^3-1\right)\left(x^3+1\right)+x^2+x+1\)
\(=\left(x^5+x^2\right)\left(x^3-1\right)+\left(x^4+x\right)\left(x^3-1\right)+x^2+x+1\)
\(=\left(x^3-1\right)\left(x^5+x^4+x^2+x\right)+x^2+x+1\)
\(=\left(x^2+x+1\right)\left(x-1\right)\left(x^5+x^4+x^2+x\right)+x^2+x+1\)
\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x\right)+x^2+x+1\)
\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)
c, \(x^4+5x^3-12x^2+5x+1\)
\(=x^4-x^3+6x^3-6x^2-6x^2+6x-x+1\)
\(=x^3\left(x-1\right)+6x^2\left(x-1\right)-6x\left(x-1\right)-\left(x-1\right)\)
\(=\left(x-1\right)\left[x^3+6x^2-6x-1\right]\)
\(=\left(x-1\right)\left[\left(x-1\right)\left(x^2+x+1\right)+6x\left(x-1\right)\right]\)
\(=\left(x-1\right)\left(x-1\right)\left(x^2+7x+1\right)\)
\(=\left(x-1\right)^2.\left(x^2+7x+1\right)\)
a, \(\left(x^2+x-2\right)\left(x^2+9x+18\right)-28\)
\(=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)-28\)
\(=\left[\left(x-1\right)\left(x+6\right)\right].\left[\left(x+2\right)\left(x+3\right)\right]-28\)
\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)-28\)
\(=\left(x^2+5x\right)^2-36-28\)
\(=\left(x^2+5x\right)^2-64\)
\(=\left(x^2+5x-8\right)\left(x^2+5x+8\right)\)
b, \(B=\left(x+1\right)^2\left(2x+3\right)-18\)
\(=\left(4x^2+8x+3\right)\left(x^2+2x+1\right)-18\)
Đặt \(x^2+2x+1=t\Rightarrow4x^2+8x+3=4t-1\)
Ta có: \(B=\left(4t-1\right)t-18\)
\(=4t^2-t-18\)
\(=4t^2-9t+8t-18\)
\(=t\left(4t-9\right)+2\left(4t-9\right)\)
\(=\left(4t-9\right)\left(t+2\right)\)
\(=\left(4x^2+8x-5\right)\left(x^2+2x+3\right)\) (vì \(t=x^2+2x+1\)
\(=\left(2x-1\right)\left(2x+5\right)\left(x^2+2x+3\right)\)
Chúc bạn học tốt.
a: \(=\dfrac{x^5}{3x}+\dfrac{12x^2}{3x}+\dfrac{5x}{3x}=\dfrac{1}{3}x^4+4x+\dfrac{5}{3}\)
b: \(=\dfrac{-5x^4}{-5x}-\dfrac{15x^3}{5x}+\dfrac{18x}{5x}=x^3-3x^2+\dfrac{18}{5}\)
c: \(=\dfrac{-x^6}{0.5x}+\dfrac{5x^4}{0.5x}-\dfrac{2x^3}{0.5x}=-2x^5+10x^3-4x^2\)
Với \(x=0\) không phải nghiệm
Với \(x\ne0\) chia 2 vế cho \(x^2\) ta được:
\(x^2-5x-12-\dfrac{5}{x}+\dfrac{1}{x^2}=0\)
\(\Leftrightarrow\left(x^2+\dfrac{1}{x^2}+2\right)-5\left(x+\dfrac{1}{x}\right)-14=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)^2-5\left(x+\dfrac{1}{x}\right)-14=0\)
Đặt \(x+\dfrac{1}{x}=t\)
\(\Rightarrow t^2-5t-14=0\Rightarrow\left[{}\begin{matrix}t=7\\t=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=-2\\x+\dfrac{1}{x}=7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+2x+1=0\\x^2-7x+1=0\end{matrix}\right.\) (bấm máy)
a/ Chắc là bạn ghi nhầm đề? Số cuối là số 9 mới đúng, chứ 27 thì câu này vô nghiệm
\(x^4+4x^3+4x^2+8x^2+12x+27=0\)
\(\Leftrightarrow x^2\left(x+2\right)^2+8\left(x+\frac{3}{4}\right)^2+\frac{45}{2}=0\)
Vế phải dương nên pt vô nghiệm
b/ Nhận thấy \(x=0\) không phải nghiệm, chia 2 vế cho \(x^2\) ta được:
\(x^2+\frac{1}{x^2}-5\left(x-\frac{1}{x}\right)+6=0\)
Đặt \(x-\frac{1}{x}=a\Rightarrow x^2+\frac{1}{x^2}=a^2+2\)
\(\Rightarrow a^2+2-5a+6=0\)
\(\Leftrightarrow a^2-5a+8=0\Rightarrow\) pt vô nghiệm
Lại nhầm đề nữa???? Dấu thứ 2 là dấu + thì pt này có nghiệm đẹp
18 - 12x = 5x - (3 - 4)
=> 18 - 12x = 5x + 1
=> 18 - 1 = 5x + 12x
=> 17 = 17x
=> x = 17 : 17
=> x = 1
18 - 12x = 5x - (3-4)
18 - 12x = 5x - 3 +4
-12x -5x = -3 +4 -18
-17x = -17
x = 1
=.= hk tốt!!