`a,(x+3)(x^2+2021)=0`

`x^2+2021>=2021>0`

`=>x+3=0`

`=>x=-3`

`2,x(x-3)+3(x-3)=0`

`=>(x-3)(x+3)=0`

`=>x=+-3`

`b,x^2-9+(x+3)(3-2x)=0`

`=>(x-3)(x+3)+(x+3)(3-2x)=0`

`=>(x+3)(-x)=0`

`=>` $\left[ \begin{array}{l}x=0\\x=-3\end{array} \right.$

`d,3x^2+3x=0`

`=>3x(x+1)=0`

`=>` $\left[ \begin{array}{l}x=0\\x=-1\end{array} \right.$

`e,x^2-4x+4=4`

`=>x^2-4x=0`

`=>x(x-4)=0`

`=>` $\left[ \begin{array}{l}x=0\\x=4\end{array} \right.$

1) a) \(\left(x+3\right).\left(x^2+2021\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^2+2021=0\end{matrix}\right.\\\left[{}\begin{matrix}x=-3\left(nhận\right)\\x^2=-2021\left(loại\right)\end{matrix}\right. \)

=> S={-3}

Bài 1: 

a) Ta có: \(\left(x+3\right)\left(x^2+2021\right)=0\)

mà \(x^2+2021>0\forall x\)

nên x+3=0

hay x=-3

Vậy: S={-3}

Bài 2: 

b) Ta có: \(x\left(x-3\right)+3\left(x-3\right)=0\)

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

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

Vậy: S={3;-3}

a: =>|x-7|=3-2x

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(-2x+3\right)^2-\left(x-7\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(2x-3-x+7\right)\left(2x-3+x-7\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(x+4\right)\left(3x-10\right)=0\end{matrix}\right.\Leftrightarrow x=-4\)

b: =>|2x-3|=4x+9

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(4x+9-2x+3\right)\left(4x+9+2x-3\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(2x+12\right)\left(6x+6\right)=0\end{matrix}\right.\Leftrightarrow x=-1\)

c: =>3x+5=2-5x hoặc 3x+5=5x-2

=>8x=-3 hoặc -2x=-7

=>x=-3/8 hoặc x=7/2

bạn tải ảnh về r up lại đi bạn

\(a,4\left(x-3\right)^2-\left(2x-1\right)^2\ge12\)

\(\Leftrightarrow4x^2-24x+36-4x^2-4x+1\ge12\)

\(\Leftrightarrow-28x+37\ge12\)

\(\Leftrightarrow-28x\ge12-37\)

\(\Leftrightarrow-28x\ge-25\)

\(\Leftrightarrow x\le\dfrac{25}{28}\)

Vậy \(S=\left\{x\left|x\le\dfrac{25}{28}\right|\right\}\)

b, \(\left(x-4\right)\left(x+4\right)\ge\left(x+3\right)^2+5\)

\(\Leftrightarrow x^2-16\ge x^2+6x+9+5\)

\(\Leftrightarrow x^2-x^2-6x\ge9+5+16\)

\(\Leftrightarrow-6x\ge30\)

\(\Leftrightarrow x\le-5\)

Vậy \(S=\left\{x\left|x\le-5\right|\right\}\)

\(c,\left(3x-1\right)^2-9\left(x+2\right)\left(x-2\right)< 5x\)

\(\Leftrightarrow9x^2-6x-1-9x^2+36< 5x\)

\(\Leftrightarrow9x^2-9x^2-6x-5x+36+1< 0\)

\(\Leftrightarrow-11x+37< 0\)

\(\Leftrightarrow-11x< -37\)

\(\Leftrightarrow x>\dfrac{37}{11}\)

vậy \(S=\left\{x\left|x>\dfrac{37}{11}\right|\right\}\)

a) đặt \(\left(x^2+x\right)\)là \(y\)

ta có: \(3y^2-7y+4\)\(=0\)

<=>\(\left(3y-4\right)\left(y-1\right)=0\)

còn lại bạn tự xử nhé 

Đề là \(\left(2x^2-x\right)^2+...\) hay là \(\left(2x^2-x\right)+...\) vậy bn?

Đặt \(2x^2-x=a\)

\(PT\Leftrightarrow a^2-9a+18=0\\ \Leftrightarrow a^2-3a-6a+18=0\\ \Leftrightarrow\left(a-3\right)\left(a-6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x^2-x-3=0\\2x^2-x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(2x-3\right)\left(x+1\right)=0\\\left(x-2\right)\left(2x+3\right)=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-1\\x=2\\x=-\dfrac{3}{2}\end{matrix}\right.\)

\(\left(2x^2-x\right)+9\left(2x^2-x\right)+18=0\)

⇔\(\left(2x^2-x\right)\left(1+9\right)+18=0\)

⇔\(10\left(2x^2-x\right)+18=0\)

⇔\(10\left(2x^2-x\right)=-18\)

⇔\(2x^2-x=-\dfrac{9}{5}\)

⇔\(x\left(2x-1\right)=-\dfrac{9}{5}\)

⇔\(x=-\dfrac{9}{5}\) hay \(2x-1=-\dfrac{9}{5}\)

⇔\(x=-\dfrac{9}{5}\) hay \(2x=-\dfrac{4}{5}\)

⇔\(x=-\dfrac{9}{5}\) hay \(x=-\dfrac{2}{5}\)

a) đkxđ \(x\ge1\)

pt đã cho \(\Leftrightarrow\left(\sqrt{2x-1}-3\right)+\left(\sqrt{x-1}-2\right)=0\)

\(\Leftrightarrow\dfrac{2x-10}{\sqrt{2x-1}+3}+\dfrac{x-5}{\sqrt{x-1}+2}=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=5\left(nhận\right)\\\dfrac{2}{\sqrt{2x-1}+3}+\dfrac{1}{\sqrt{x-1}+3}=0\end{matrix}\right.\)

 Hiển nhiên pt thứ 2 vô nghiệm vì \(VT>0\) với mọi \(x\ge1\). Do đó pt đã cho có nghiệm duy nhất là \(x=5\)

b) đkxđ: \(x\ge-3\)

 Để ý rằng \(x^2+2x+7=\left(x^2+1\right)+\left(2x+6\right)=\left(x^2+1\right)+2\left(x+3\right)\) nên nếu ta đặt \(\sqrt{x^2+1}=u\left(u\ge1\right)\) và \(\sqrt{x+3}=v\left(v\ge0\right)\) thì pt đã chot rở thành:

 \(u^2+2v^2=3uv\)

 \(\Leftrightarrow\left(u-v\right)\left(u-2v\right)=0\)

 \(\Leftrightarrow\left[{}\begin{matrix}u=v\\u=2v\end{matrix}\right.\)

Nếu \(u=v\) thì \(\sqrt{x^2+1}=\sqrt{x+3}\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge-3\\x^2+1=x+3\end{matrix}\right.\) 

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

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

 Nếu \(u=2v\) thì \(\sqrt{x^2+1}=2\sqrt{x+3}\)

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

mà \(x^2+1=4x+12\)\(\Leftrightarrow x^2-4x-11=0\)

\(\Leftrightarrow x=2\pm\sqrt{15}\) (nhận)

Vậy pt đã cho có tập nghiệm \(S=\left\{2;-1;2\pm\sqrt{15}\right\}\)

a) \(\sqrt{2x-1}+\sqrt{x-1}=5\) (ĐK: \(x\ge1\)) 

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

\(\Leftrightarrow2x-1+x-1+2\sqrt{\left(2x-1\right)\left(x-1\right)}=25\)

\(\Leftrightarrow3x-2+2\sqrt{\left(2x-1\right)\left(x-1\right)}=25\)

\(\Leftrightarrow\sqrt{\left(2x-1\right)\left(x-1\right)}=\dfrac{27-3x}{2}\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{27-3x}{2}\ge0\\\left(2x-1\right)\left(x-1\right)=\left(\dfrac{27-3x}{2}\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}27-3x\ge0\\2x^2-2x-x+1=\dfrac{729-162x+9x^2}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x\le27\\8x^2-12x+4=9x^2-162x+729\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le9\\x^2-150x+725=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le9\\\left[{}\begin{matrix}x-5=0\\x-145=0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le9\\\left[{}\begin{matrix}x=5\left(tm\right)\\x=145\left(ktm\right)\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow x=5\)

7,3, -6

ĐKXĐ: \(x\ne7;x\ne2\)

BPT \(\Leftrightarrow f\left(x\right)=\dfrac{\left(6-2x\right)^3\left(x+6\right)}{\left(x-7\right)^3}\le0\)

Lập bảng xét dấu ta có:

Từ đây ta thấy \(-6\le x\le3\) hoặc \(x>7\) thỏa mãn bất phương trình ban đầu.

Vậy...

a) 4         b) 0,6 hoặc 1,75     e) -3,5 hoặc -3