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\(ĐKXĐ:x\ne0\)
\(8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)^2-4\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)^2=\left(x+4\right)^2\)\(\Leftrightarrow8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)\left(x^2+\dfrac{1}{x^2}-\left(x+\dfrac{1}{x}\right)^2\right)=\left(x+4\right)^2\)\(\Leftrightarrow8\left(x+\dfrac{1}{x}\right)^2-8\left(x^2+\dfrac{1}{x^2}\right)=\left(x+4\right)^2\)
\(\Leftrightarrow16=\left(x+4\right)^2\Leftrightarrow\)\(\left[{}\begin{matrix}x=-8\\x=0\end{matrix}\right.\) \(\Rightarrow x=-8\) (vì \(x\ne0\))
\(S=\left\{-8\right\}\)
Đặt \(x+\dfrac{1}{x}=a\)
ta có \(\left(x+\dfrac{1}{x}\right)^2=a^2\Rightarrow x^2+2+\dfrac{1}{x^2}=a^2\Rightarrow x^2+\dfrac{1}{x^2}=a^2-2\)
ta có \(8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)^2-4\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)^2=\left(x+4\right)^2\)\(\Leftrightarrow8a^2+4.\left(a^2-2\right)^2-4\left(a^2-2\right)a^2=\left(x+4\right)^2\)
\(\Leftrightarrow8a^2+4\left(a^4-4a^2+4\right)-4a^4+8a^2=\left(x+4\right)^2\)
\(\Leftrightarrow8a^2+4a^4-16a^2+16-4a^4+8a^2-\left(x+4\right)^2=0\)
\(\Leftrightarrow\left(x+4\right)^2=16\)
\(\Leftrightarrow x+4=4\) hoặc \(x+4=-4\)
\(\Leftrightarrow x=-4\) ( thỏa mãn x\(\ne\)0) hoặc x=0 (ktm x\(\ne\)0)
vậy x=-4
Điều kiện \(x\ne0\)
\(\Leftrightarrow8.\dfrac{x^4+2x^2+1}{x^2}+4.\dfrac{x^8+2x^4+1}{x^4}-4.\dfrac{x^4+1}{x^2}.\dfrac{x^4+2x^2+1}{x^2}=\left(x^2+8x+16\right)\)
\(\Leftrightarrow x^2+8x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(l\right)\\x=-8\end{matrix}\right.\)
Hung nguyen,Ace Legona và những ai có thể giải bài này,help me!!
2.a)
\(2x\left(6x-1\right)>\left(3x-2\right)\left(4x+3\right)\)
\(\Leftrightarrow12x^2-2x>12x^2+9x-8x-6\)
\(\Leftrightarrow12x^2-2x-12x^2-9x+8x>6\)
\(\Leftrightarrow-3x>6\)
\(\Leftrightarrow3>\dfrac{6}{-3}\)
\(\Leftrightarrow x< -2\)
Vậy nghiệm của bpt \(S=\left\{-2\right\}\)
2.b)
\(\dfrac{2\left(x+1\right)}{3}-2\ge\dfrac{x-2}{2}\)
\(\Leftrightarrow4\left(x+1\right)-2.6\ge3x-6\)
\(\Leftrightarrow4x+4-12\ge3x-6\)
\(\Leftrightarrow4x-3x\ge-6-4+12\)
\(\Leftrightarrow x\ge2\)
vậy nghiệm của bpt x\(\ge\)2
\(8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)^2-4\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)^2=\left(x+4\right)^2\) ⇔ \(8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)\left(x^2+\dfrac{1}{x^2}-x^2-\dfrac{1}{x^2}-2\right)=\left(x+4\right)^2\) ⇔ \(8\left(x+\dfrac{1}{x}\right)^2-8\left(x^2+\dfrac{1}{x^2}\right)=\left(x+4\right)^2\) ( x # 0 )
⇔ \(8\left(x^2+\dfrac{1}{x^2}+2-x^2-\dfrac{1}{x^2}\right)=\left(x+4\right)^2\)
⇔ \(x^2+8x=0\)
⇔ \(x=0\left(KTM\right)orx=-8\left(TM\right)\)
KL...............
a ) \(\dfrac{1}{x+1}-\dfrac{5}{x-2}=\dfrac{15}{\left(x+1\right)\left(2-x\right)}\)(1)
ĐKXĐ : \(x\ne1;x\ne2\)
(1)\(\Leftrightarrow\dfrac{1}{x+1}+\dfrac{5}{2-x}=\dfrac{15}{\left(x+1\right)\left(2-x\right)}\)
\(\Leftrightarrow2-x+5x+5=15\)
\(\Leftrightarrow4x+7=15\\\)
\(\Leftrightarrow4x=8\)
\(\Leftrightarrow x=2\left(KTMĐKXĐ\right)\)
Vậy pt vô nghiệm .
b ) \(1+\dfrac{x}{3-x}=\dfrac{5x}{\left(x+2\right)\left(3-x\right)}+\dfrac{2}{x+2}\) ( 2 )
ĐKXĐ : \(x\ne3;x\ne-2\)
(2) \(\Leftrightarrow3x-x^2+6-2x+x^2+2x=3x+6-x^2-2x\)
\(\Leftrightarrow x^2+2x=0\)
\(\Leftrightarrow x\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(TMĐKXĐ\right)\\x=-2\left(KTMĐKXĐ\right)\end{matrix}\right.\)
Vậy tập nghiệm của phương trình là S={0}.
c ) \(\dfrac{6}{x-1}-\dfrac{4}{x-3}=\dfrac{8}{\left(x-1\right)\left(3-x\right)}\) (3)
ĐKXĐ : \(x\ne1;x\ne3\)
\(\left(3\right)\Leftrightarrow\dfrac{6}{x-1}+\dfrac{4}{3-x}=\dfrac{8}{\left(x-1\right)\left(3-x\right)}\)
\(\Leftrightarrow6\left(3-x\right)+4\left(x-1\right)=8\)
\(\Leftrightarrow18-6x+4x-4=8\)
\(\Leftrightarrow-2x=6\)
\(\Leftrightarrow x=-3\)
Vậy tập nghiệm của phương trình là S={-3}
d ) \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\) (4)
ĐKXĐ : \(x\ne0;x\ne2\)
\(\left(4\right)\Leftrightarrow x^2+2x-x+2=2\)
\(\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(KTMĐKXĐ\right)\\x=-1\left(TMĐKXĐ\right)\end{matrix}\right.\)
Vậy tập nghiệm của phương trình là S={-1}
a) \(\dfrac{1}{x+1}-\dfrac{5}{x-2}=\dfrac{15}{\left(x+1\right)\left(2-x\right)}\) ( đk: x ≠ -1; x ≠ 2 )
\(\Leftrightarrow\) \(\dfrac{1}{x+1}+\dfrac{5}{2-x}=\dfrac{15}{\left(x+1\right)\left(2-x\right)}\)
\(\Leftrightarrow\) \(2-x+5\left(x+1\right)=15\)
\(\Leftrightarrow\) \(2-x+5x+5=15\)
\(\Leftrightarrow\)\(4x=8\)
\(\Rightarrow\) \(x=2\) ( KTM )
S = ∅
b) \(1+\dfrac{x}{3-x}=\dfrac{5x}{\left(x+2\right)\left(3-x\right)}+\dfrac{2}{x+2}\) ( đk: x ≠ - 2 ; x ≠ 3 )
\(\Leftrightarrow\) \(\left(x+2\right)\left(3-x\right)+x\left(x+2\right)=5x+2\left(3-x\right)\)
\(\Leftrightarrow\) \(3x-x^2+6-2x+x^2+2x=5x+6-2x\)
\(\Leftrightarrow\) \(3x+6=3x+6\)
\(\Rightarrow\)\(0x=0\) ( TM )
\(\Rightarrow\) Phương trình vô số nghiệm
S = R
c) \(\dfrac{6}{x-1}-\dfrac{4}{x-3}=\dfrac{8}{\left(x-1\right)\left(3-x\right)}\) ( đk: x ≠ 1 ; x ≠ 3 )
\(\Leftrightarrow\) \(\dfrac{6}{x-1}+\dfrac{4}{3-x}=\dfrac{8}{\left(x-1\right)\left(3-x\right)}\)
\(\Leftrightarrow\)\(6\left(3-x\right)+4\left(x-1\right)=8\)
\(\Leftrightarrow\) \(18-6x+4x-4=8\)
\(\Leftrightarrow\) \(-2x=-6\)
\(\Rightarrow x=3\) ( KTM )
S = ∅
d) \(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x\left(x-2\right)}\) (đk: x ≠ 2; x ≠ 0 )
\(\Leftrightarrow\) \(x\left(x+2\right)-x+2=2\)
\(\Leftrightarrow\) \(x^2+2x-x+2=2\)
\(\Leftrightarrow\) \(x^2+x=0\)
\(\Leftrightarrow\) \(x\left(x+1\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=0\left(KTM\right)\\x=1\left(TM\right)\end{matrix}\right.\)
S = \(\left\{2\right\}\)
đkxđ với mọi x
đặt a=x2+x+1
\(\dfrac{a}{a+1}+\dfrac{a+1}{a+2}=\dfrac{7}{6}\)
<=> \(\dfrac{6a\left(a+2\right)}{6\left(a+1\right)\left(a+2\right)}+\dfrac{6\left(a+1\right)^2}{6\left(a+1\right)\left(a+2\right)}=\dfrac{7\left(a+1\right)\left(a+2\right)}{6\left(a+1\right)\left(a+2\right)}\)
=> 6a(a+2) +6(a+1)2 =7(a+1)(a+2)
<=> 6a2+12a +6a2 +12a+6 =a2 +21a+14
<=> 12a2 -a2+24a-21a+6-14=0
<=> 11a2+3a-8=0
<=> 11a2 +11a-8a-8=0
<=> (11a2 +11a)-(8a+8)=0
<=> 11a(a+1)-8(a+1)=0
<=> (a+1)(11a-8)=0
=> a=-1 và a=\(\dfrac{8}{11}\)
thay a=x2+x+1 ta đc
x2+x+1=-1
<=> x2+x+2 =0 (vô nghiệm)
và x2+x+\(\dfrac{3}{11}\) =0(vô nghiệm )
vậy pt trên vô nghiệm
c) \(8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)^2-4\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)^2=\left(x+4\right)^2\left(2\right)\)ĐKXĐ : x # 0
( 2) <=> \(8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)\left[\left(x^2+\dfrac{1}{x^2}\right)-\left(x+\dfrac{1}{x}\right)^2\right]=\left(x+4\right)^2\)
\(< =>8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right).\left(-2\right)=\left(x+4\right)^2\)
\(< =>8.\left[\left(x+\dfrac{1}{x}\right)^2-x^2-\dfrac{1}{x^2}\right]=\left(x+4\right)^2\)
\(< =>16=\left(x+4\right)^2\)
<=> x2 + 8x = 0
<=> x( x + 8) = 0
<=> x = 0 ( KTM ) hoặc x = - 8 ( TM )
Vậy,....
Bài 17)
(x - 2)^4 + (x - 6)^4 = 82
Đặt t = x + 3
=> x + 2 = t - 1; x + 4 = t + 1.
ta có pt: (t - 1)^4 + (t + 1)^4 = 82
<=>[(t -1)²]² + [(t + 1)²]² = 82
<=> (t² - 2t + 1)² + (t² + 2t + 1)² = 82
<=> (t²+1)² - 4t(t²+1) + 4t² + (t²+1)² + 4t(t²+1) + 4t² = 82
<=> (t² + 1)² + 4t² = 41
<=> t^4 + 6t² + 1 = 41
<=> (t²)² + 6t² - 40 = 0
<=> t² = -10 (loại) hoặc t² = 4
<=> t = 2 hoặc t = -2
với t = -2 => x = -5
với t = 2 => x = -1
vậy pt có hai nghiệm là : x = -1 hoặc x = -5
Bài 18: Phương trình đã cho được viết thành: $${({x^2} + 6x + 10)^2} + (x + 3)\left[ {3\left( {{x^2} + 6x + 10} \right) + 2\left( {x + 3} \right)} \right] = 0$$
Đặt $u = {x^2} + 6x + 10 > 0,v = x + 3$, suy ra:
$${u^2} + v\left( {3u + 2v} \right) = 0 \Leftrightarrow \left( {u + v} \right)\left( {u + 2v} \right) = 0 \Leftrightarrow \left[ \begin{gathered}
u + v = 0 \\
u + 2v = 0 \\
\end{gathered} \right.$$
$$ \Leftrightarrow \left[ \begin{gathered}
{x^2} + 6x + 10 + x + 3 = 0 \\
{x^2} + 6x + 10 + 2\left( {x + 3} \right) = 0 \\
\end{gathered} \right. \Leftrightarrow \left[ \begin{gathered}
{x^2} + 7x + 13 = 0 \\
{x^2} + 8x + 16 = 0 \\
\end{gathered} \right. \Leftrightarrow x = - 4$$
Ta có: \(8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)^2-4\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)^2=\left(x+4\right)^2\)