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Thay x = 1 vào phương trình 2(2x+1)+18=3(x+2)(2x+k)2(2x+1)+18=3(x+2)(2x+k), ta có:
2(2.1+1)+18=3(1+2)(2.1+k)
⇔2(2+1)+18=3.3(2+k)⇔2.3+18=9(2+k)
⇔6+18=18+9k⇔24−18=9k⇔6=9k
⇔k=69=232(2.1+1)+18=3(1+2)(2.1+k)
⇔2(2+1)+18=3.3(2+k)
⇔2.3+18=9(2+k)
⇔6+18=18+9k
⇔24−18=9k⇔6=9k
⇔k=\(\frac{6}{9}\)=\(\frac{2}{3}\)
Vậy khi thì phương trình có nghiệm x = 1
a. Thay x = 2 vào phương trình (2x + 1)(9x + 2k) – 5(x + 2) = 40, ta có:
(2.2+1)(9.2+2k)−5(2+2)=40⇔(4+1)(18+2k)−5.4=40⇔5(18+2k)−20=40⇔90+10k−20=40⇔10k=40−90+20⇔10k=−30⇔k=−3(2.2+1)(9.2+2k)−5(2+2)=40⇔(4+1)(18+2k)−5.4=40⇔5(18+2k)−20=40⇔90+10k−20=40⇔10k=40−90+20⇔10k=−30⇔k=−3
Vậy khi k = -3 thì phương trình (2x + 1)(9x + 2k) – 5(x + 2) = 40 có nghiệm x = 2
b. Thay x = 1 vào phương trình 2(2x+1)+18=3(x+2)(2x+k)2(2x+1)+18=3(x+2)(2x+k), ta có:
2(2.1+1)+18=3(1+2)(2.1+k)⇔2(2+1)+18=3.3(2+k)⇔2.3+18=9(2+k)⇔6+18=18+9k⇔24−18=9k⇔6=9k⇔k=69=232(2.1+1)+18=3(1+2)(2.1+k)⇔2(2+1)+18=3.3(2+k)⇔2.3+18=9(2+k)⇔6+18=18+9k⇔24−18=9k⇔6=9k⇔k=\(\frac{6}{9}\)=\(\frac{2}{3}\)
Vậy khi thì phương trình có nghiệm x = 1
Đặt x2 + 2x = a ta có
\(\frac{1}{a-3}\)+ \(\frac{18}{a+2}\)= \(\frac{18}{a+1}\)
<=> a2 - 15a + 56 = 0
<=> a = (7;8)
Thế vô tìm được nghiệm
\(1,\left(x+y\right)^2-\left(x-y\right)^2=\left[\left(x+y\right)-\left(x-y\right)\right]\left[\left(x+y\right)+\left(x-y\right)\right]=\left(x+y-x+y\right)\left(x+y+x-y\right)=2y.2x=4xy\)
\(2,\left(x+y\right)^3-\left(x-y\right)^3-2y^3\)
\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3\)
\(=6x^2y\)
\(3,\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\\ =\left[\left(x+y\right)-\left(x-y\right)\right]^2\\ =\left(x+y-x+y\right)^2\\ =4y^2\)
\(4,\left(2x+3\right)^2-2\left(2x+3\right)\left(2x+5\right)+\left(2x+5\right)^2\\ =\left[\left(2x+3\right)-\left(2x+5\right)\right]^2\\ =\left(2x+3-2x-5\right)^2\\ =\left(-2\right)^2\\ =4\)
\(5,9^8.2^8-\left(18^4+1\right)\left(18^4-1\right)\\ =18^8-\left[\left(18^4\right)^2-1\right]\\ =18^8-18^8+1\\ =1\)
1: =x^2+2xy+y^2-x^2+2xy-y^2=4xy
2: =x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-2y^3
=6x^2y
3: =(x+y-x+y)^2=(2y)^2=4y^2
4: =(2x+3-2x-5)^2=(-2)^2=4
5: =18^8-18^8+1=1
\(\dfrac{11x}{2x-3}+\dfrac{x-18}{2x-3}\left(ĐKXĐ:x\ne\dfrac{3}{2}\right)\\ =\dfrac{11x+x-18}{2x-3}\\ =\dfrac{12x-18}{2x-3}\\ =\dfrac{6\left(2x-3\right)}{2x-3}\\ =6\)
\(\dfrac{2x+12}{4x^2-9}+\dfrac{2x+5}{4x-6}\left(ĐKXĐ:x\ne\dfrac{3}{2};x\ne\dfrac{-3}{2}\right)\\ =\dfrac{2x+12}{\left(2x-3\right)\left(2x+3\right)}+\dfrac{2x+5}{2\left(2x-3\right)}\\ =\dfrac{4x+24}{2\left(2x-3\right)\left(2x+3\right)}+\dfrac{\left(2x+5\right)\left(2x+3\right)}{2\left(2x-3\right)\left(2x+3\right)}\\ =\dfrac{4x+24+4x^2+6x+10x+15}{2\left(2x-3\right)\left(2x+3\right)}\\ =\dfrac{4x^2+20x+39}{2\left(2x-3\right)\left(2x+3\right)}\)
\(\dfrac{x}{2x+1}+\dfrac{-1}{4x^2-1}+\dfrac{2-x}{2x-1}\left(ĐKXĐ:x\ne\dfrac{1}{2};x\ne\dfrac{-1}{2}\right)\\ =\dfrac{x\left(2x-1\right)-1+\left(2-x\right)\left(2x+1\right)}{\left(2x+1\right)\left(2x-1\right)}\\ =\dfrac{2x^2-x-1+4x+2-2x^2-x}{\left(2x-1\right)\left(2x+1\right)}\\ =\dfrac{2x+1}{\left(2x+1\right)\left(2x-1\right)}\\ =\dfrac{1}{2x-1}\)
\(a,=\dfrac{4x+8}{x^2+2x}=\dfrac{4\left(x+2\right)}{x\left(x+2\right)}=\dfrac{4}{x}\\ b,=\dfrac{\left(2x-3\right)-\left(2x-4\right)}{x-2}=\dfrac{2x-3-2x+4}{x-2}=\dfrac{1}{x-2}\\ c,=\dfrac{2x-1-3x-2}{x+3}=\dfrac{-x-3}{x+3}=\dfrac{-\left(x+3\right)}{x+3}=-1\\ d,=\dfrac{11x-18+x}{2x-3}=\dfrac{12x-18}{2x-3}=\dfrac{6\left(2x-3\right)}{2x-3}=6\)
\(e,=\dfrac{3x-6-9x+3}{2x+1}=\dfrac{-6x-3}{2x+1}=\dfrac{-3\left(2x+1\right)}{2x+1}=-3\)
\(\dfrac{1}{x^2+2x-3}+\dfrac{18}{x^2+2x+2}=\dfrac{18}{x^2+2x+1}\left(1\right)\)
ĐK: \(x\ne\pm1,x\ne-3\)
Đặt \(y=x^2+2x+1\) (với y > 0,y khác 4) ta được:
\(\left(1\right)\Leftrightarrow\dfrac{1}{y-4}+\dfrac{18}{y+1}=\dfrac{18}{y}\Leftrightarrow\dfrac{y\left(y+1\right)}{y\left(y+1\right)\left(y-4\right)}+\dfrac{18y\left(y-4\right)}{y\left(y+1\right)\left(y-4\right)}=\dfrac{18\left(y+1\right)\left(y-4\right)}{y\left(y+1\right)\left(y-4\right)}\Rightarrow y\left(y+1\right)+18y\left(y-4\right)=18\left(y+1\right)\left(y-4\right)\Leftrightarrow y^2+y+18y^2-72y=18y^2-54y-72\Leftrightarrow y^2-17y+72=0\Leftrightarrow\left(y-8\right)\left(y-9\right)=0\Leftrightarrow\left[{}\begin{matrix}y=8\left(TM\right)\\y=9\left(TM\right)\end{matrix}\right.\)
Với \(y=8\) ta có :
\(x^2+2x+1=8\Leftrightarrow\left(x+1\right)^2=8\Leftrightarrow x+1=\pm\sqrt{8}\Leftrightarrow x=\pm\sqrt{8}-1\)
Với y=9 ta có:
\(x^2+2x+1=9\Leftrightarrow\left(x+1\right)^2=9\Leftrightarrow x+1=\pm\sqrt{9}\Leftrightarrow x=\pm\sqrt{9}-1\)
ĐKXĐ : \(x\ne1;-3\)
Đặt \(x^2+2x+1=a\) , ta có :
\(\frac{1}{a-4}+\frac{18}{a+1}=\frac{18}{a}\)
\(\Leftrightarrow\frac{a+1+18a-72}{\left(a+1\right)\left(a-4\right)}=\frac{18}{a}\)
\(\Leftrightarrow\frac{19a-71}{a^2-3a-4}=\frac{18}{a}\)
\(\Leftrightarrow19a^2-71a-18a^2+54a+72=0\)
\(\Leftrightarrow a^2-17a+72=0\)
\(\Leftrightarrow\left(a-8\right)\left(a-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=8\\a=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(x+1\right)^2=8\\\left(x+1\right)^2=9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\pm\sqrt{8}-1\\\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\end{matrix}\right.\)
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