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1:
\(=\left(\dfrac{1}{x-2\sqrt{x}}+\dfrac{2}{3\sqrt{x}-6}\right):\dfrac{2\sqrt{x}+3}{3\sqrt{x}}\)
\(=\dfrac{3+2\sqrt{x}}{3\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{3\sqrt{x}}{2\sqrt{x}+3}=\dfrac{1}{\sqrt{x}-2}\)
a) (3x2 - 7x – 10)[2x2 + (1 - √5)x + √5 – 3] = 0
=> hoặc (3x2 - 7x – 10) = 0 (1)
hoặc 2x2 + (1 - √5)x + √5 – 3 = 0 (2)
Giải (1): phương trình a - b + c = 3 + 7 - 10 = 0
nên
x1 = - 1, x2 = =
Giải (2): phương trình có a + b + c = 2 + (1 - √5) + √5 - 3 = 0
nên
x3 = 1, x4 =
b) x3 + 3x2– 2x – 6 = 0 ⇔ x2(x + 3) – 2(x + 3) = 0 ⇔ (x + 3)(x2 - 2) = 0
=> hoặc x + 3 = 0
hoặc x2 - 2 = 0
Giải ra x1 = -3, x2 = -√2, x3 = √2
c) (x2 - 1)(0,6x + 1) = 0,6x2 + x ⇔ (0,6x + 1)(x2 – x – 1) = 0
=> hoặc 0,6x + 1 = 0 (1)
hoặc x2 – x – 1 = 0 (2)
(1) ⇔ 0,6x + 1 = 0
⇔ x2 = =
(2): ∆ = (-1)2 – 4 . 1 . (-1) = 1 + 4 = 5, √∆ = √5
x3 = , x4 =
Vậy phương trình có ba nghiệm:
x1 = , x2 =
, x3 =
,
d) (x2 + 2x – 5)2 = ( x2 – x + 5)2 ⇔ (x2 + 2x – 5)2 - ( x2 – x + 5)2 = 0
⇔ (x2 + 2x – 5 + x2 – x + 5)( x2 + 2x – 5 - x2 + x - 5) = 0
⇔ (2x2 + x)(3x – 10) = 0
⇔ x(2x + 1)(3x – 10) = 0
Hoặc x = 0, x = , x =
Vậy phương trình có 3 nghiệm:
x1 = 0, x2 = , x3 =
b: Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=0\)
\(\Leftrightarrow\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24=0\)
\(\Leftrightarrow\left(x^2+7x\right)^2+22\left(x^2+7x\right)+120-24=0\)
\(\Leftrightarrow x^2+7x+6=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\)
a: \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+2\right)\left(y+3\right)=xy+100\\\left(x-2\right)\left(y-2\right)=xy-64\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x+2y=94\\-2x-2y=-68\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=26\\y=8\end{matrix}\right.\)
b: \(\Leftrightarrow\left\{{}\begin{matrix}-3x+2y=0\\-x+y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=3\end{matrix}\right.\)
c: \(\Leftrightarrow\left\{{}\begin{matrix}xy-2x=xy-4x+2y-8\\2xy+7x-6y-21=2xy+6x-7y-21\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-2y=-8\\x+y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-2\\y=2\end{matrix}\right.\)
\(a.\left(\sqrt{x}-7\right)\left(\sqrt{x}-8\right)=x+11\left(x\ge0\right)\)
\(\Leftrightarrow x-15\sqrt{x}+56=x+11\)
\(\Leftrightarrow15\sqrt{x}=45\)
\(\Leftrightarrow x=9\left(TM\right)\)
\(b.\left(\sqrt{x}+3\right)\left(\sqrt{x}-5\right)=x-17\left(x\ge0\right)\)
\(\Leftrightarrow x-2\sqrt{x}-15=x-17\)
\(\Leftrightarrow2\sqrt{x}=2\)
\(x=1\left(TM\right)\)
\(c.1-\dfrac{2\sqrt{x}-5}{6}=\dfrac{3-\sqrt{x}}{4}\left(x\ge0\right)\)
\(\Leftrightarrow\dfrac{2\left(2\sqrt{x}-5\right)+3\left(3-\sqrt{x}\right)}{12}=1\)
\(\Leftrightarrow x=169\left(TM\right)\)
\(d.\left(\sqrt{x}+3\right)^2-x+3=0\left(x\ge0\right)\)
\(\Leftrightarrow6\sqrt{x}=-12\left(vô-lý\right)\)
KL...............
Câu a :
\(x-5\sqrt{x}-14=0\)
\(\Leftrightarrow\left(\sqrt{x}+2\right)\left(\sqrt{x}-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+2=0\\\sqrt{x}-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\\x=49\end{matrix}\right.\)
Vậy \(S=\left\{49\right\}\)
Câu b :
\(\left(x^2+x+1\right)\left(x^2+x+2\right)=2\)
Đặt \(x^2+x+1=t\)
\(\Leftrightarrow t\left(t+1\right)=2\)
\(\Leftrightarrow t^2+t-2=0\)
\(\Leftrightarrow\left(t-1\right)\left(t+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t-1=0\\t+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=1\\t=-2\end{matrix}\right.\)
Với \(t=1\) thì :
\(x^2+x+1=1\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Với \(t=-2\) thì :
\(x^2+x+1=-2\)
\(\Leftrightarrow x^2+x+3=0\) ( pt vô nghiệm )
Vậy \(S=\left\{-1;0\right\}\)