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\(x^6-x^5+x^4-x^3+x^2-x+\dfrac{3}{4}=0\)
\(\Rightarrow\left(x^6-x^5\right)+\left(x^4-x^3\right)+\left(x^2-x\right)+\dfrac{3}{4}=0\)
\(\Rightarrow x^5\left(x-1\right)+x^3\left(x-1\right)+x\left(x-1\right)+\dfrac{3}{4}=0\)
\(\Rightarrow\left(x-1\right)\left(x^5+x^3+x\right)+\dfrac{3}{4}=0\)
.... bí cmnr :))
\(\frac{-5}{9}x+1=\frac{2}{3}x-10\)
\(\frac{-5}{9}x+\frac{9}{9}=\frac{6}{9}x-\frac{90}{9}\)
\(-5x+9=6x-90\)
\(-5x-6x=-90-9\)
\(-11x=-99\)
\(x=\frac{-99}{-11}=9\)
b. \(\frac{x-22}{8}+\frac{x-21}{9}+\frac{x-20}{10}+\frac{x-19}{11}=4\)
\(\frac{x-22}{8}-1+\frac{x-21}{9}-1+\frac{x-20}{10}-1+\frac{x-19}{11}-1=0\)
\(\frac{x-30}{8}+\frac{x-30}{9}+\frac{x-30}{10}+\frac{x-30}{11}=0\)
\(\left(x-30\right)\left(\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{11}\right)=0\)
x=30
Chúc bạn học tốt!!
a) \(\left(2x+1\right)\left(3x-2\right)=\left(2x+1\right)\left(5x-8\right)\)
\(\Leftrightarrow\)\(\left(2x+1\right)\left(3x-2\right)-\left(2x+1\right)\left(5x-8\right)=0\)
\(\Leftrightarrow\)\(\left(2x+1\right)\left(3x-2-5x+8\right)=0\)
\(\Leftrightarrow\)\(\left(2x+1\right)\left(6-2x\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}2x+1=0\\6-2x=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-0,5\\x=3\end{cases}}\)
Vậy...
b) \(ĐKXĐ:\) \(x\ne-2;\) \(x\ne4\)
\(\frac{3}{x+2}+\frac{2}{x-4}=0\)
\(\Leftrightarrow\)\(\frac{3\left(x-4\right)}{\left(x+2\right)\left(x-4\right)}+\frac{2\left(x+2\right)}{\left(x+2\right)\left(x-4\right)}=0\)
\(\Leftrightarrow\)\(\frac{3x-12+2x+4}{\left(x+2\right)\left(x-4\right)}=0\)
\(\Leftrightarrow\)\(\frac{5x-8}{\left(x+2\right)\left(x-4\right)}=0\)
\(\Rightarrow\)\(5x-8=0\)
\(\Leftrightarrow\)\(x=\frac{8}{5}\) (T/m đkxđ)
Vậy...
c) \(x^3+4x^2+4x+3=0\)
\(\Leftrightarrow\)\(x^3+3x^2+x^2+3x+x+3=0\)
\(\Leftrightarrow\)\(x^2\left(x+3\right)+x\left(x+3\right)+\left(x+3\right)=0\)
\(\Leftrightarrow\)\(\left(x+3\right)\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\)\(x+3=0\) (do \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\) \(\forall x\))
\(\Leftrightarrow\)\(x=-3\)
Vậy...
Nhận thấy \(x=0\) ko phải nghiệm, chia 2 vế cho \(x^2\)
\(x^2-3x+9-\frac{3}{x}+\frac{1}{x^2}=0\)
\(\Leftrightarrow x^2+\frac{1}{x^2}-3\left(x+\frac{1}{x}\right)+9=0\)
Đặt \(x+\frac{1}{x}=t\Rightarrow x^2+\frac{1}{x^2}=t^2-2\)
pt trở thành: \(t^2-2-3t+9=0\)
\(\Leftrightarrow t^2-3t+7=0\) (vô nghiệm)
Vậy pt đã cho vô nghiệm
a) 2 (x + 3) (x + 4) - (x - 2)2 > (x - 1)2
<=> 2 (x2 + 4x + 3x + 12) - (x2 - 4x + 4) > x2 - 2x + 1
<=> 2x2 + 8x + 6x + 24 - x2 + 4x - 4 - x2 + 2x - 1 > 0
<=> 20x + 20 > 0
<=> 20 (x + 1) > 0
=> x + 1 > 0
=> x > -1
b) 5x2 - 18x + 19 - (2x - 3)2 > 0
<=> 5x2 - 18x + 19 - (2x2 - 12x + 9) > 0
<=> 5x2 - 18x + 19 - 2x2 + 12x - 9 > 0
<=> 3x2 - 6x + 10 > 0
\(\Delta=b^2-4ac=\left(-6\right)^2-4.3.10=-84< 0\)
Vậy pt vô nghiệm hay\(x\in\varnothing\)
#Học tốt!!!
a) 2( x + 3 )( x + 4 ) - ( x - 2 )2 > ( x - 1 )2
<=> 2( x2 + 7x + 12 ) - ( x2 - 4x + 4 ) > x2 - 2x + 1
<=> 2x2 + 14x + 24 - x2 + 4x - 4 - x2 + 2x - 1 > 0
<=> 20x + 19 > 0
<=> x > -19/20
Vậy ...
\(5X\left(X-2020\right)+X=2020\)
\(\Leftrightarrow5X^2-10100X+X=2020\)
\(\Leftrightarrow5X^2-10099X=2020\)
\(\Leftrightarrow5X^2-10099X-2020=0\)
\(\Leftrightarrow5X^2-10100X+x-2020=0\)
\(\Leftrightarrow5X\left(X-2020\right)+X-2020=0\)
\(\Leftrightarrow\left(X-2020\right)\left(5X+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2020\\x=-\frac{1}{5}\end{cases}}\)
\(4\left(x-5\right)^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left[2\left(x-5\right)\right]^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left[2\left(x-5\right)-2x-1\right]\left[2\left(x-5\right)+2x+1\right]=0\)
\(\Leftrightarrow\left(2x-10-2x-1\right)\left(2x-10+2x+1\right)=0\)
\(\Leftrightarrow-11\left(4x-9\right)=0\)
\(\Leftrightarrow x=\frac{9}{4}\)
Pt da cho <=> (x+3)(x-3-4(x+3))=0<=>(x+3)(-3x-15)=0 <=> x+3=0 hoặc -3x-15=0 <=> x=-3 hoặc x= -5