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\(a,\left(2x+1\right)^2-3x^2+4=\left(1-x\right)\left(1+x\right)\)
\(\Leftrightarrow4x^2+4x+1-3x^2+4=1-x^2\)
\(\Leftrightarrow4x^2+4x+1-3x^2+4-1+x^2=0\)
\(\Leftrightarrow2x^2+4x+4=0\)
\(\Leftrightarrow2\left(x^2+2x+1\right)+2=0\)
\(\Leftrightarrow2\left(x+1\right)^2=-2\)
\(\Leftrightarrow\left(x+1\right)^2=-1\Rightarrow\) pt vô nghiệm
\(b,\left(4x-3\right)\left(4x+3\right)-2\left(x+2\right)^2=14x^2\)
\(\Leftrightarrow16x^2-9-2\left(x^2+4x+4\right)-14x^2=0\)
\(\Leftrightarrow16x^2-9-2x^2-8x-8-14x^2=0\)
\(\Leftrightarrow-8x-17=0\)
\(\Leftrightarrow-8x=17\)
\(\Leftrightarrow x=\dfrac{-17}{8}\)
\(c,\left(2x-1\right)\left(x+1\right)-x^2+1=\dfrac{1}{2}\left(x-1\right)^2\)
\(\Leftrightarrow2x^2+2x-x-1-x^2+1=\dfrac{1}{2}\left(x^2-2x+1\right)\)
\(\Leftrightarrow2x^2+2x-x-1-x^2+1-\dfrac{1}{2}x^2+x-\dfrac{1}{2}=0\)\(\Leftrightarrow\dfrac{1}{2}x^2+2x-\dfrac{1}{2}=0\)
\(\Leftrightarrow\dfrac{1}{2}\left(x^2+4x+4\right)-\dfrac{5}{2}=0\)
\(\Leftrightarrow\dfrac{1}{2}\left(x+2\right)^2=\dfrac{5}{2}\)
\(\Rightarrow\left(x+2\right)^2=5\)
\(\Rightarrow\left[{}\begin{matrix}x+2=-\sqrt{5}\\x+2=\sqrt{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\sqrt{5}-2\\x=\sqrt{5}-2\end{matrix}\right.\)
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 \(x=-2+\sqrt{5};x=-2-\sqrt{5}\)
1) điều kiện xác định : \(x\notin\left\{-1;-2;-3;-4\right\}\)
ta có : \(\dfrac{1}{x^2+3x+2}+\dfrac{1}{x^2+5x+6}+\dfrac{1}{x^2+7x+12}=\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\) \(\Leftrightarrow\dfrac{\left(x+3\right)\left(x+4\right)+\left(x+1\right)\left(x+4\right)+\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)\(\Leftrightarrow\dfrac{x^2+7x+12+x^2+5x+4+x^2+3x+2}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{3x^2+15x+18}{\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)}=\dfrac{1}{6}\)
\(\Leftrightarrow6\left(3x^2+15x+18\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(\Leftrightarrow18\left(x^2+5x+6\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(\Leftrightarrow18\left(x+2\right)\left(x+3\right)=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\)
\(\Leftrightarrow18=\left(x+1\right)\left(x+4\right)\) ( vì điều kiện xác định )
\(\Leftrightarrow18=x^2+5x+4\Leftrightarrow x^2+5x-14=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+7=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-7\end{matrix}\right.\left(tmđk\right)\)
vậy \(x=2\) hoặc \(x=-7\) mấy câu kia lm tương tự nha bn
1: Ta có: \(4x^2-36=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
2: Ta có: \(\left(x-1\right)^2+x\left(4-x\right)=11\)
\(\Leftrightarrow x^2-2x+1+4x-x^2=11\)
\(\Leftrightarrow2x=10\)
hay x=5
a: Ta có: \(\left(7x+4\right)^2-\left(7x-4\right)\left(7x+4\right)\)
\(=\left(7x+4\right)\left(7x+4-7x+4\right)\)
\(=8\left(7x+4\right)\)
=56x+32
b: Ta có: \(8\left(x-2\right)^2-3\left(x^2-4x-5\right)-5x^2\)
\(=8x^2-32x+32-3x^2+12x+15-5x^2\)
\(=-20x+47\)
c: Ta có: \(\left(x+1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3x\left(x+1\right)\)
\(=x^3+3x^2+3x+1-x^3+1-3x^2-3x\)
=2
\(M=\left(7-2x\right)\left(4x^2+14x+49\right)-\left(64-8x^3\right)\)
\(M=\left(7-2x\right)\left[\left(2x\right)^2+2x\cdot7+7^2\right]-\left(64-8x^3\right)\)
\(M=\left[7^3-\left(2x\right)^3\right]-\left(64-8x^3\right)\)
\(M=343-8x^3-64+8x^3\)
\(M=279\)
Vậy M có giá trị 279 với mọi x
\(P=\left(2x-1\right)\left(4x^2-2x+1\right)-\left(1-2x\right)\left(1+2x+4x^2\right)\)
\(P=8x^3-4x^2+2x-4x^2+2x-1-1+8x^3\)
\(P=16x^3-8x^2+4x-2\)
Thay \(x=10\) vào P ta có:
\(P=16\cdot10^3-8\cdot10^2+4\cdot10-2=15238\)
Vậy P có giá trị 15238 tại x=10
a: M=343-8x^3-64+8x^3=279
b: P=8x^3-4x^2+2x-4x^2+2x-1-1+8x^3
=16x^3-8x^2+4x-2
=16*10^3-8*10^2+4*10-2=15238
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:
$${u^2} + v\left( {3u + 2v} \right) = 0 \Leftrightarrow \left( {u + v} \right)\left( {u + 2v} \right) = 0 \Leftrightarrow \left[ \begin{gathered}
u + v = 0 \\
u + 2v = 0 \\
\end{gathered} \right.$$
$$ \Leftrightarrow \left[ \begin{gathered}
{x^2} + 6x + 10 + x + 3 = 0 \\
{x^2} + 6x + 10 + 2\left( {x + 3} \right) = 0 \\
\end{gathered} \right. \Leftrightarrow \left[ \begin{gathered}
{x^2} + 7x + 13 = 0 \\
{x^2} + 8x + 16 = 0 \\
\end{gathered} \right. \Leftrightarrow x = - 4$$
\(\left(x^2-8\right)^2-36\)
\(=\left(x^2-8-6\right)\left(x^2-8+6\right)\)
\(=\left(x^2-14\right)\left(x^2-2\right)\)
\(\left(x^2-8\right)^2-36\)
\(=\left(x^2-8\right)^2-6^2\)
\(=\left(x^2-14\right)\left(x^2-2\right)\)