a) Ta có: \(x^2+3x-10=0\)

\(\Leftrightarrow x^2+5x-2x-10=0\)

\(\Leftrightarrow x\left(x+5\right)-2\left(x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

Vậy: S={-5;2}

b) Ta có: \(3x^2-7x+1=0\)

\(\Leftrightarrow3\left(x^2-\dfrac{7}{3}x+\dfrac{1}{3}\right)=0\)

mà 3>0

nên \(x^2-\dfrac{7}{3}x+\dfrac{1}{3}=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}-\dfrac{37}{36}=0\)

\(\Leftrightarrow\left(x-\dfrac{7}{6}\right)^2=\dfrac{37}{36}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{7}{6}=\dfrac{\sqrt{37}}{6}\\x-\dfrac{7}{6}=-\dfrac{\sqrt{37}}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{37}+7}{6}\\x=\dfrac{-\sqrt{37}+7}{6}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{\sqrt{37}+7}{6};\dfrac{-\sqrt{37}+7}{6}\right\}\)

c) Ta có: \(3x^2-7x+8=0\)

\(\Leftrightarrow3\left(x^2-\dfrac{7}{3}x+\dfrac{8}{3}\right)=0\)

mà 3>0

nên \(x^2-\dfrac{7}{3}x+\dfrac{8}{3}=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}+\dfrac{47}{36}=0\)

\(\Leftrightarrow\left(x-\dfrac{7}{6}\right)^2=-\dfrac{47}{36}\)(vô lý)

Vậy: \(x\in\varnothing\)

ko bt

c.

ĐLXĐ: \(x\ge-\dfrac{1}{3}\)

\(-\left(3x+1\right)+\sqrt{3x+1}+4x^2-10x+6=0\)

Đặt \(\sqrt{3x+1}=t\ge0\)

\(\Rightarrow-t^2+t+4x^2-10x+6=0\)

\(\Delta=1+4\left(4x^2-10x+6\right)=\left(4x-5\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{-1+4x-5}{-2}=3-2x\\t=\dfrac{-1-4x+5}{-2}=2x-2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3x+1}=3-2x\left(x\le\dfrac{3}{2}\right)\\\sqrt{3x-1}=2x-2\left(x\ge1\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+1=4x^2-12x+9\left(x\le\dfrac{3}{2}\right)\\3x-1=4x^2-8x+4\left(x\ge1\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)

a.

ĐKXĐ: \(x\ge-\dfrac{5}{4}\)

\(\Leftrightarrow4x^2-12x-2-2\sqrt{4x+5}=0\)

\(\Leftrightarrow\left(4x^2-8x+4\right)-\left(4x+5+2\sqrt{4x+5}+1\right)=0\)

\(\Leftrightarrow\left(2x-2\right)^2-\left(\sqrt{4x+5}+1\right)^2=0\)

\(\Leftrightarrow\left(2x-2-\sqrt{4x+5}-1\right)\left(2x-2+\sqrt{4x+5}+1\right)=0\)

\(\Leftrightarrow\left(2x-3-\sqrt{4x+5}\right)\left(2x-1+\sqrt{4x+5}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{4x+5}=2x-3\left(x\ge\dfrac{3}{2}\right)\\\sqrt{4x+5}=1-2x\left(x\le\dfrac{1}{2}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+5=4x^2-12x+9\left(x\ge\dfrac{3}{2}\right)\\4x+5=4x^2-4x+1\left(x\le\dfrac{1}{2}\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)

a) Ta có: \(3x^2+2x-1=0\)

\(\Leftrightarrow3x^2+3x-x-1=0\)

\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{-1;\dfrac{1}{3}\right\}\)

b) Ta có: \(x^2-5x+6=0\)

\(\Leftrightarrow x^2-2x-3x+6=0\)

\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

Vậy: S={2;3}

c) Ta có: \(x^2-3x+2=0\)

\(\Leftrightarrow x^2-x-2x+2=0\)

\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

Vậy: S={1;2}

d) Ta có: \(2x^2-6x+1=0\)

\(\Leftrightarrow2\left(x^2-3x+\dfrac{1}{3}\right)=0\)

mà \(2\ne0\)

nên \(x^2-3x+\dfrac{1}{3}=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{23}{12}=0\)

\(\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2=\dfrac{23}{12}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{2}=\dfrac{\sqrt{69}}{6}\\x-\dfrac{3}{2}=\dfrac{-\sqrt{69}}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9+\sqrt{69}}{6}\\x=\dfrac{9-\sqrt{69}}{6}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{9+\sqrt{69}}{6};\dfrac{9-\sqrt{69}}{6}\right\}\)

e) Ta có: \(4x^2-12x+5=0\)

\(\Leftrightarrow4x^2-10x-2x+5=0\)

\(\Leftrightarrow2x\left(2x-5\right)-\left(2x-5\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-5=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=5\\2x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{5}{2};\dfrac{1}{2}\right\}\)

cho vào máy tính là ra hết

         (12x-1)(6x-1)(4x-1)(3x-1)=5

<=>(12x-1)(12x-2)(12x-3)(12x-4)=40

<=>[(12x-1)(12x-4)] [(12x-2)(12x-3)] =40

<=>(144x^2 - 60x + 4) (144x^2 - 60x + 6) =40

đặt 144x^2 - 60x +4 = t  =>144x^2 - 60x +6 = t+2

ta có phương trình:

        t ( t+2 ) =40

<=> t^2 + 2t -40 =0

<=> (t+1)^2 -39 =0

<=> t+1=\(\sqrt{39}\)      hoặc t+1=\(-\sqrt{39}\)    <=> x=\(\sqrt{39}\) -1 hoặc x=\(-\sqrt{39}\) -1

tick nha

cách 1: phân tích ra ước

cách 2 áp dụng 7 hằng đẳng thức nhân tung ra

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