\(8,1-\left(x-6\right)=4\left(2-2x\right)\)

\(\Leftrightarrow1-x+6=8-8x\)

\(\Leftrightarrow-x+8x=8-1-6\)

\(\Leftrightarrow7x=1\)

\(\Leftrightarrow x=\dfrac{1}{7}\)

\(9,\left(3x-2\right)\left(x+5\right)=0\)

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

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

\(10,\left(x+3\right)\left(x^2+2\right)=0\)

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

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

`8)1-(x-5)=4(2-2x)`

`<=>1-x+5=8-6x`

`<=>5x=2<=>x=2/5`

`9)(3x-2)(x+5)=0`

`<=>[(x=2/3),(x=-5):}`

`10)(x+3)(x^2+2)=0`

  Mà `x^2+2 > 0 AA x`

 `=>x+3=0`

`<=>x=-3`

`11)(5x-1)(x^2-9)=0`

`<=>(5x-1)(x-3)(x+3)=0`

`<=>[(x=1/5),(x=3),(x=-3):}`

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

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

`<=>[(x=3),(x=-3):}`

`13)x(x-5)-4x+20=0`

`<=>x(x-5)-4(x-5)=0`

`<=>(x-5)(x-4)=0`

`<=>[(x=5),(x=4):}`

`14)x^2+4x-5=0`

`<=>x^2+5x-x-5=0`

`<=>(x+5)(x-1)=0`

`<=>[(x=-5),(x=1):}`

\(11,=>\left[{}\begin{matrix}5x-1=0\\x^2-9=0\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=3\\x=-3\end{matrix}\right.\\ 12,=>\left(x+3\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x+3=0\\x-3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\\ 13,=>x\left(x-5\right)-4\left(x-5\right)=0\\ =>\left(x-4\right)\left(x-5\right)=0\\ =>\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

\(14,=>x^2+5x-x-5=0\\ =>x\left(x+5\right)-\left(x+5\right)=0\\ =>\left(x-1\right)\left(x+5\right)=0\\ =>\left[{}\begin{matrix}x-1=0\\x+5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)

len google di ban

mk chua hoc bai nay

\(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2+3x^2=-33\)

<=> \(x^3-9x^2+27x-27\) \(-\left(x^3-3^3\right)+6\left(x^2+2x+1\right)+3x^2=-33\)

<=> \(x^3-9x^2+27x-27-x^3+27+6x^2+12x+6+3x^2=-33\)

<=> \(-6x^2+39x+6=-33\)

<=> \(6x^2-39x-6=33\)

<=> \(6x^2-39x-39=0\)

<=> \(6\left(x^2-\frac{39}{6}x-\frac{39}{6}\right)=0\)

<=> \(x^2-2.x.\frac{39}{12}+\frac{1521}{144}-\frac{273}{16}=0\)

<=> \(\left(x-\frac{39}{12}\right)^2-\frac{273}{16}=0\)

<=> \(\left(x-\frac{39}{12}-\frac{\sqrt{273}}{4}\right)\left(x-\frac{39}{12}+\frac{\sqrt{273}}{4}\right)=0\)

<=> \(\left(x-\frac{13+\sqrt{273}}{4}\right).\left(x-\frac{13-\sqrt{273}}{4}\right)=0\)

<=> \(x=\frac{13+\sqrt{273}}{4}\) ( h ) \(x=\frac{13-\sqrt{273}}{4}\)

học tốt

\(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2+3x^2=-33\)

\(\Leftrightarrow\left(x-3\right)^3-\left(x-3\right)^3+6\left(x+1\right)^2+3x^2=-33\)

\(\Leftrightarrow6\left(x+1\right)^2+3x^2=-33\text{ vô lý }\left(\text{vì }6\left(x+1\right)^2\ge0;3x^2\ge0\right)\)

\(\text{Vậy không có x nào thỏa mãn}\)

Rảnh rỗi thật sự .-.

a: =>|3x-5|=|x+2|

=>3x-5=x+2 hoặc 3x-5=-x-2

=>2x=7 hoặc 4x=3

=>x=7/2 hoặc x=3/4

b: \(\Leftrightarrow\left\{{}\begin{matrix}3x-5=0\\x+2=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

c: \(\Leftrightarrow\left|3x-5\right|=x-2\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=2\\\left(3x-5-x+2\right)\left(3x-5+x-2\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=2\\\left(2x-3\right)\left(4x-7\right)=0\end{matrix}\right.\Leftrightarrow x\in\varnothing\)

d: \(\dfrac{11}{2}\le\left|x\right|< \dfrac{17}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{11}{2}< =x< \dfrac{17}{2}\\-\dfrac{17}{2}< x< =-\dfrac{11}{2}\end{matrix}\right.\)

(x-1)^2+(y-1)^2=25

=>R=5; I(1;1)

\(d\left(I;\text{Δ}\right)=\dfrac{\left|1\cdot3+1\cdot4+33\right|}{\sqrt{3^2+4^2}}=\dfrac{40}{5}=8>5\)

=>Δ nằm ngoài (C)

Lập đường thẳng đi qua I và vuông góc với 3x+4y+33=0

=>(d'): -4x+3y+c=0

Thay x=1 và y=1 vào (d'), ta được:

c-4+3=0

=>c=1

=>-4x+3y+1=0

-4x+3y+1=0 và (x-1)^2+(y-1)^2=25

=>-4x=-3y-1 và (x-1)^2+(y-1)^2=25

=>x=3/4y+1/4 và (3/4y+1/4-1)^2+(y-1)^2=25

=>9/16(y-1)^2+(y-1)^2=25 và x=3/4y+1/4

=>(y-1)^2=16 và x=3/4y+1/4

=>(y=5 hoặc y=-3) và x=3/4y+1/4

=>(x,y)=(4;5) hoặc (x,y)=(-2;-3)

=>M1(4;5); M2(-2;-3)

Δ: 3x+4y+33=0; (d'): -4x+3y+1=0

=>H(-19/5; -27/5)

\(M_1H=\sqrt{\left(-\dfrac{19}{5}-4\right)^2+\left(-\dfrac{27}{5}-5\right)^2}=13\)

\(M_2H=\sqrt{\left(-\dfrac{19}{5}+2\right)^2+\left(-\dfrac{27}{5}+3\right)^2}=3\)

=>\(d_{min}=3;d_{max}=13\)

\(c,\Rightarrow\left[{}\begin{matrix}-2\left(x+2\right)+\left(4-x\right)=11\left(x< -2\right)\\2\left(x+2\right)+\left(4-x\right)=11\left(-2\le x\le4\right)\\2\left(x+2\right)+\left(x-4\right)=11\left(x>4\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{11}{3}\left(tm\right)\\x=3\left(tm\right)\\x=\dfrac{11}{3}\left(ktm\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{11}{3}\end{matrix}\right.\)

\(a,\Rightarrow\left[{}\begin{matrix}x+\dfrac{5}{2}=3x+1\\x+\dfrac{5}{2}=-3x-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{7}{8}\end{matrix}\right.\)

\(b,\Rightarrow\left[{}\begin{matrix}6-2x-x-3=8\left(x\le-3\right)\\6-2x+x+3=8\left(-3\le x\le3\right)\\2x-6+x+3=8\left(x>3\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{-5}{3}\left(ktm\right)\\x=1\left(tm\right)\\x=\dfrac{11}{3}\left(tm\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{11}{3}\end{matrix}\right.\)