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( x +1 ) ( x + 4 ) = 5 căn ( x^2 + 5x +28 ) (1)
= ( x + 1 ) ( x + 4 ) = 5 căn [ (x^2 + 5x + 4) + 24 ]
= ( x + 1 ) ( x + 4 ) = 5 căn [ ( x + 1 ) ( x + 4 ) + 24 ]
Đặt a = ( x + 1 ) ( x + 4 )
(1) <=> a = 5 căn ( a + 24 )
<=> a^2 = 25 ( a + 24 )
<=> a^2 - 25a - 600 = 0
<=> a1 = 40
a2 = -15
với a = 40 ta có:
( x + 1 ) ( x + 4 ) = 40
<=> x^2 + 5x + 4 = 40
<=> x^2 + 5x - 36 = 0
<=> x = 4 và x = - 9
với a = -15, ta có:
( x + 1 ) ( x + 4 ) = -15
<=> x^2 + 5x + 4 = -15
<=> x^2 + 5x + 19 = 0
delta < 0 => pt vô nghiệm
Vậy s = { -9; 4}
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2\sqrt{3x-2}-2=11x+\sqrt{5x+6}-3\sqrt{\left(3x-2\right)\left(5x+6\right)}\)
ĐK: \(x\ge\dfrac{2}{3}\)
\(pt\Leftrightarrow2\sqrt{3x-2}-2-11x-\sqrt{5x+6}+3\sqrt{\left(3x-2\right)\left(5x+6\right)}=0\)
\(\Leftrightarrow2\sqrt{3x-2}-4-11x+22-\sqrt{5x+6}+4+3\sqrt{\left(3x-2\right)\left(5x+6\right)}-24=0\)
\(\Leftrightarrow2\dfrac{3x-2-16}{\sqrt{3x-2}+4}-11\left(x-2\right)-\dfrac{5x+6-16}{\sqrt{5x+6}+4}+\dfrac{9\left(3x-2\right)\left(5x+6\right)-576}{3\sqrt{\left(3x-2\right)\left(5x+6\right)}+24}=0\)
\(\Leftrightarrow\dfrac{6\left(x-2\right)}{\sqrt{3x-2}+4}-11\left(x-2\right)-\dfrac{5\left(x-2\right)}{\sqrt{5x+6}+4}+\dfrac{9\left(x-2\right)\left(15x+38\right)}{3\sqrt{\left(3x-2\right)\left(5x+6\right)}+24}=0\)
\(\Leftrightarrow\left(x-2\right)\left(\dfrac{6}{\sqrt{3x-2}+4}-11-\dfrac{5}{\sqrt{5x+6}+4}+\dfrac{9\left(15x+38\right)}{3\sqrt{\left(3x-2\right)\left(5x+6\right)}+24}\right)=0\)
\(\Rightarrow x-2=0\Rightarrow x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1. 3x( x - 2 ) - ( x - 2 ) = 0
<=> ( x-2).(3x-1) = 0 => x = 2 hoặc x = \(\dfrac{1}{3}\)
2. x( x-1 ) ( x2 + x + 1 ) - 4( x - 1 )
<=> ( x - 1 ).( x (x^2 + x + 1 ) - 4 ) = 0
(phần này tui giải được x = 1 thôi còn bên kia giải ko ra nha )
3 \(\left\{{}\begin{matrix}\sqrt{5}x-2y=7\\\sqrt{5}x-5y=10\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}y=-1\\x=\sqrt{5}\end{matrix}\right.\)
\(1. 3x^2 - 7x +2=0\)
=>\(Δ=(-7)^2 - 4.3.2\)
\(= 49-24 = 25\)
Vì 25>0 suy ra phương trình có 2 nghiệm phân biệt:
\(x_1\)=\(\dfrac{-\left(-7\right)+\sqrt{25}}{2.3}=\dfrac{7+5}{6}=2\)
\(x_2\)=\(\dfrac{-\left(-7\right)-\sqrt{25}}{2.3}=\dfrac{7-5}{6}=\dfrac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(x^2+5x+3=t\) (1) \(\Rightarrow x^2+5x=t-3\)
Pt: \(\Rightarrow\)\(x^2+5x+6-2\sqrt{t}=6\)
\(\Rightarrow t-3-6-2\sqrt{t}-6=0\Rightarrow t-2\sqrt{t}-15=0\)
\(\Rightarrow\left[{}\begin{matrix}t=5\\t=-3\end{matrix}\right.\)
Thay t=5 vào (1) ta đc: \(x^2+5x+3=5\Rightarrow x^2+5x-2=0\Rightarrow x=\dfrac{-5\pm\sqrt{33}}{2}\)
Thay t=-3 vào (1) ta đc:
\(x^2+5x+3=-3\Rightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
`a,3x^2+7x+2=0`
`<=>3x^2+6x+x+2=0`
`<=>3x(x+2)+x+2=0`
`<=>(x+2)(3x+1)=0`
`<=>x=-2\or\x=-1/3`
d) Ta có: (x-1)(x+2)=70
\(\Leftrightarrow x^2+2x-x-2-70=0\)
\(\Leftrightarrow x^2+x-72=0\)
\(\Leftrightarrow x^2+9x-8x-72=0\)
\(\Leftrightarrow x\left(x+9\right)-8\left(x+9\right)=0\)
\(\Leftrightarrow\left(x+9\right)\left(x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+9=0\\x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=8\end{matrix}\right.\)
Vậy: S={8;-9}
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2\left(x^2+3\right)-\left(7x+1\right)\sqrt{x^2+3}+3x^2+3x=0\)
Đặt \(\sqrt{x^2+3}=t>0\)
\(\Rightarrow2t^2-\left(7x+1\right)t+3x^2+3x=0\)
\(\Delta=\left(7x+1\right)^2-8\left(3x^2+3x\right)=25x^2-10x+1=\left(5x-1\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}t=\frac{7x+1-\left(5x-1\right)}{4}=\frac{x+1}{2}\\t=\frac{7x+1+5x-1}{4}=3x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+3}=\frac{x+1}{2}\left(x\ge-1\right)\\\sqrt{x^2+3}=3x\left(x\ge0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+3=\frac{x^2+2x+1}{4}\\x^2+3=9x^2\end{matrix}\right.\) \(\Leftrightarrow...\)
Đặt \(t=\sqrt{\left(x^2+3\right)}\)
\(\Rightarrow t^2=x^2+3\)
\(\Rightarrow\)Phương trình trở thành
\(7t^2-\left(11x-1\right)t-5\left(x+3\right)=0\)
Delta = \(\left(11x-1\right)+4.7.5\left(x+3\right)>0\forall x\)
'-' Đến đây bạn tìm nghiệm t theo ẩn x sau đó thay \(t=\sqrt{\left(x^2+3\right)}\)để tìm ra đáp án nhé !