cm gtbt:P=\(\dfrac{\left(x^2+a\right)\cdot\left(1+a\right)+a^2\cdot x^2+1}{\left(x^2-a\right)\cdot\left(1-a\right)+a^2\cdot x^2+1}\) không phụ thuộc vào x
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8 tháng 7 2017
\(a,2\left(5x+1\right)-7\left(3x-2\right)=4\left(2x-1\right)+3\left(2-x\right)\)
\(\Leftrightarrow10x+2-21x+14=8x-4+6-3x\)
\(\Leftrightarrow-16x=-14\)
\(\Rightarrow x=\dfrac{7}{8}\)
\(b,-4\left(\dfrac{1}{2}x-3\right)+\dfrac{7}{2}\left(2x-1\right)+x=5x\left(1-x\right)\)
\(\Leftrightarrow-2x+12+7x-\dfrac{7}{2}+x=5x-5x^2\)
\(\Leftrightarrow5x^2+x+\dfrac{17}{2}=0\)
Cái này không biết tách kiểu gì cho vừa nên bạn nhấn máy tính nhé
Mode 5 3 rồi lần lượt điền vào theo thứ tự trên thì
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{10}+\dfrac{13i}{10}\\x=-\dfrac{1}{10}-\dfrac{13i}{10}\end{matrix}\right.\)
KL
3 tháng 8 2017
1. a, 3x + |x - 2| = 8
<=> |x - 2| = 8 - 3x
Xét 2 TH :
TH1: x - 2 = 8 - 3x
<=> x + 3x = 8 + 2
<=> 4x = 10
<=> x = \(\dfrac{5}{2}\) (thỏa mãn)
TH2: x - 2 = -(8 - 3x)
<=> x - 2 = -8 + 3x
<=> -2 + 8 = 3x - x
<=> 6 = 2x
<=> x = 3 (thỏa mãn)
b, 5 - |x - 1| = 4
<=> |x - 1| = 1
<=> \(\left[{}\begin{matrix}x-1=1\\x-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\) (thỏa mãn)
@Nguyễn Hoàng Vũ
KL
3 tháng 8 2017
2. 5.(x - 2) - 4.(1 - 3x) = |3 - 7| + 2.(1 + 2x)
<=> 5x - 10 - 4 + 12x = 4 + 2 + 4x
<=> 17x - 14 = 6 + 4x
<=> 17x - 4x = 6 + 14
<=> 13x = 20
<=> x = \(\dfrac{20}{13}\) (thỏa mãn)
@Nguyễn Hoàng Vũ
TN
17 tháng 6 2019
\(A=\left(a^2+b^2-c^2\right)^2-\left(a^2-b^2+c^2\right)^2-4a^2b^2\)
\(=\left(a^2+b^2-c^2+a^2-b^2+c^2\right)\left(a^2+b^2-c^2-a^2+b^2-c^2\right)-4a^2b^2\)
\(=2a^2.2b^2-4a^2b^2=0\)
\(C=\left(2-6x\right)^2+\left(2-5x\right)^2+2\left(6x-2\right)\left(2-5x\right)\)
\(=\left[\left(2-6x\right)+\left(2-5x\right)\right]^2\)
\(=\left[4-11x\right]^2\)
\(=16-88x+121x^2\)
chúc bn học tốt
PN
13 tháng 7 2017
Bài 1:
\(\left(-\dfrac{72}{40}-\dfrac{144}{60}-2\dfrac{1}{3}\right):\left(\dfrac{45}{100}-\dfrac{25}{60}+-\dfrac{75}{25}\right)\)
\(=\left(-\dfrac{9}{5}-\dfrac{12}{5}-\dfrac{7}{3}\right):\left(\dfrac{9}{20}-\dfrac{5}{12}+-3\right)\)
\(=\left(-\dfrac{27}{15}-\dfrac{36}{15}-\dfrac{21}{15}\right):\left(\dfrac{27}{60}-\dfrac{25}{60}+-3\right)\)
\(=\left(-\dfrac{28}{5}\right):\left(-\dfrac{89}{30}\right)\)
\(=\left(-\dfrac{28}{5}\right).\left(-\dfrac{30}{89}\right)\)
\(=\dfrac{168}{89}\)
TN
12 tháng 7 2017
\(a,\left(2x+1\right)^2-3x^2+4=\left(1-x\right)\left(1+x\right)\)
\(\Leftrightarrow4x^2+4x+1-3x^2+4=1-x^2\)
\(\Leftrightarrow4x^2+4x+1-3x^2+4-1+x^2=0\)
\(\Leftrightarrow2x^2+4x+4=0\)
\(\Leftrightarrow2\left(x^2+2x+1\right)+2=0\)
\(\Leftrightarrow2\left(x+1\right)^2=-2\)
\(\Leftrightarrow\left(x+1\right)^2=-1\Rightarrow\) pt vô nghiệm
\(b,\left(4x-3\right)\left(4x+3\right)-2\left(x+2\right)^2=14x^2\)
\(\Leftrightarrow16x^2-9-2\left(x^2+4x+4\right)-14x^2=0\)
\(\Leftrightarrow16x^2-9-2x^2-8x-8-14x^2=0\)
\(\Leftrightarrow-8x-17=0\)
\(\Leftrightarrow-8x=17\)
\(\Leftrightarrow x=\dfrac{-17}{8}\)
\(c,\left(2x-1\right)\left(x+1\right)-x^2+1=\dfrac{1}{2}\left(x-1\right)^2\)
\(\Leftrightarrow2x^2+2x-x-1-x^2+1=\dfrac{1}{2}\left(x^2-2x+1\right)\)
\(\Leftrightarrow2x^2+2x-x-1-x^2+1-\dfrac{1}{2}x^2+x-\dfrac{1}{2}=0\)\(\Leftrightarrow\dfrac{1}{2}x^2+2x-\dfrac{1}{2}=0\)
\(\Leftrightarrow\dfrac{1}{2}\left(x^2+4x+4\right)-\dfrac{5}{2}=0\)
\(\Leftrightarrow\dfrac{1}{2}\left(x+2\right)^2=\dfrac{5}{2}\)
\(\Rightarrow\left(x+2\right)^2=5\)
\(\Rightarrow\left[{}\begin{matrix}x+2=-\sqrt{5}\\x+2=\sqrt{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\sqrt{5}-2\\x=\sqrt{5}-2\end{matrix}\right.\)
12 tháng 7 2017
a) \(\left(2x+1\right)^2-3x^2+4=\left(1-x\right)\left(1+x\right)\)
\(\Leftrightarrow4x^2+4x+1-3x^2+4=1-x^2\)
\(\Leftrightarrow4x^2+4x+1-3x^2+4-1+x^2=0\)
\(\Leftrightarrow2x^2+4x+4=0\Leftrightarrow\left(\sqrt{2}x\right)^2+2.\sqrt{2}.\sqrt{2}x+\left(\sqrt{2}\right)^2+2=0\) \(\Leftrightarrow\left(\sqrt{2}x+\sqrt{2}\right)^2+2=0\)
ta có : \(\left(\sqrt{2}x+\sqrt{2}\right)^2\ge0\Rightarrow\left(\sqrt{2}x+\sqrt{2}\right)^2+2\ge2>0\forall x\)
\(\Rightarrow\) phương trình vô nghiệm
vậy phương trình vô nghiệm
b) \(\left(4x-3\right)\left(4x+3\right)-2\left(x+2\right)^2=14x^2\)
\(\Leftrightarrow16x^2-9-2\left(x^2+4x+4\right)=14x^2\)
\(\Leftrightarrow16x^2-9-2x^2-8x-8=14x^2\)
\(\Leftrightarrow16x^2-9-2x^2-8x-8-14x^2=0\)
\(\Leftrightarrow-8x-17=0\Leftrightarrow-8x=17\Leftrightarrow x=\dfrac{-17}{8}\)
vậy \(x=\dfrac{-17}{8}\)
c) \(\left(2x-1\right)\left(x+1\right)-x^2+1=\dfrac{1}{2}\left(x-1\right)^2\)
\(\Leftrightarrow2x^2+2x-x-1-x^2+1=\dfrac{1}{2}\left(x^2-2x+1\right)\)
\(\Leftrightarrow2x^2+2x-x-1-x^2+1=\dfrac{1}{2}x^2-x+\dfrac{1}{2}\)
\(\Leftrightarrow2x^2+2x-x-1-x^2+1-\dfrac{1}{2}x^2+x-\dfrac{1}{2}=0\)
\(\Leftrightarrow\dfrac{1}{2}x^2+2x-\dfrac{1}{2}=0\Leftrightarrow\left(\dfrac{\sqrt{2}}{2}x\right)^2+2.\sqrt{2}.\dfrac{\sqrt{2}}{2}x+\left(\sqrt{2}\right)^2-\dfrac{5}{2}=0\)
\(\Leftrightarrow\left(\dfrac{\sqrt{2}}{2}x+\sqrt{2}\right)^2-\dfrac{5}{2}=0\Leftrightarrow\left(\dfrac{\sqrt{2}}{2}x+\sqrt{2}\right)^2=\dfrac{5}{2}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{\sqrt{2}}{2}x+\sqrt{2}=\sqrt{\dfrac{5}{2}}\\\dfrac{\sqrt{2}}{2}x+\sqrt{2}=-\sqrt{\dfrac{5}{2}}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{\sqrt{2}}{2}x=\sqrt{\dfrac{5}{2}}-\sqrt{2}=\dfrac{\sqrt{10}-2\sqrt{2}}{2}\\\dfrac{\sqrt{2}}{2}x=-\sqrt{\dfrac{5}{2}}-\sqrt{2}=-\dfrac{\sqrt{10}+2\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-2+\sqrt{5}\\x=-2-\sqrt{5}\end{matrix}\right.\)
vậy \(x=-2+\sqrt{5};x=-2-\sqrt{5}\)
22 tháng 5 2022
1: \(=8x^3+12x^2+6x+1-8x^3+12x^2-6x+1-2\left(4x+3\right)^2+8\left(x+3\right)^2\)
\(=24x^2+2-2\left(16x^2+24x+9\right)+8\left(x^2+6x+9\right)\)
\(=24x^2+2-32x^2-48x-18+8x^2+48x+72\)
=56
2: \(=\left(4x^2+4x+1\right)\left(x-1\right)-2\left(x^3-6x^2+12x-8\right)+x\left(3-2x\right)\left(3+x\right)-\left(3x-3\right)^2\)
\(=4x^3-3x-1-2x^3+12x^2-24x+16+x\left(9-3x-2x^2\right)-\left(3x-3\right)^2\)
\(=2x^3+12x^2-27x+15+9x-3x^2-2x^3-9x^2+18x-9\)
\(=6\)
TN
13 tháng 7 2017
\(a,4x^2-\left(3x+1\right)\left(2x-1\right)=2\left(x-3\right)^2\)
\(\Leftrightarrow4x^2-\left(6x^2-3x+2x-1\right)=2\left(x^2-6x+9\right)\)
\(\Leftrightarrow4x^2-6x^2+x+1-2x^2+12x-18=0\)
\(\Leftrightarrow-4x^2+13x-17=0\)
\(\Leftrightarrow-4\left(x^2-\dfrac{13}{4}x+\dfrac{169}{64}\right)-\dfrac{103}{16}=0\)
\(\Leftrightarrow-4\left(x-\dfrac{13}{8}\right)^2=\dfrac{103}{16}\)
\(\Leftrightarrow\left(x-\dfrac{13}{8}\right)^2=\dfrac{-103}{64}\Rightarrow\) pt vô nghiệm
\(b,\left(5x-1\right)\left(x+1\right)-\left(2x-1\right)\left(2x+1\right)=x.\left(x+1\right)\)\(\Leftrightarrow5x^2+5x-x-1-\left(4x^2-1\right)=x^2+x\)
\(\Leftrightarrow5x^2+5x-x-1-4x^2+1-x^2-x=0\) \(\Leftrightarrow3x=0\Rightarrow x=0\)
\(c,7x^2-\left(2x-3\right)^2=1+3\left(x+2\right)^2\)
\(\Leftrightarrow7x^2-\left(4x^2-12x+9\right)=1+3\left(x^2+4x+4\right)\)
\(\Leftrightarrow7x^2-4x^2+12x-9=1+3x^2+12x+12\)\(\Leftrightarrow7x^2-4x^2+12x-9-1-3x^2-12x-12=0\)\(\Leftrightarrow-22=0\) ( vô lí)
Vậy phương trình vô nghiệm
=(x^2+x^2a+a+a^2+a^2x^2+1)/(x^2-x^2a-a+a^2+a^2x^2+1)
=\(\dfrac{x^2\left(1+a^2\right)+a^2+1+x^2a+a}{x^2+a^2x^2+a^2+1-x^2a-a}\)
\(=\dfrac{\left(1+a^2\right)\left(x^2+1\right)+a\left(x^2+1\right)}{x^2\left(1+a^2\right)+\left(a^2+1\right)-a\left(x^2+1\right)}\)
\(=\dfrac{\left(x^2+1\right)\left(a^2+a+1\right)}{\left(a^2+1\right)\left(x^2+1\right)-a\left(x^2+1\right)}=\dfrac{a^2+a+1}{a^2-a+1}\)