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

\(\left(x+2\right)\left(x+1-x+3\right)=0\Leftrightarrow x=-2\)

\(\left(x-2\right)\left(x+3\right)-\left(x-2\right)\left(2x+5\right)=0\Leftrightarrow\left(x-2\right)\left(-x-2\right)=0\Leftrightarrow x=-2;x=2\)

\(i,\left(x-1\right)\left(x+3\right)-\left(x-1\right)\left(2x+1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x+3-2x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(-x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\\ k,\left(x+2\right)\left(x+1\right)-\left(x-3\right)\left(x+2\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x+1-x+3\right)=0\\ \Leftrightarrow4\left(x+2\right)=0\\ \Leftrightarrow x+2=0\\ \Leftrightarrow x=-2\\ l,\left(x-2\right)\left(x+3\right)=\left(x-2\right)\left(2x+5\right)\\ \Leftrightarrow\left(x-2\right)\left(2x+5\right)-\left(x-2\right)\left(x+3\right)=0\\ \Leftrightarrow\left(x-2\right)\left(2x+5-x-3\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

1) Ta có: \(\left(3-x^2\right)+6-2x=0\)

\(\Leftrightarrow3-x^2+6-2x=0\)

\(\Leftrightarrow-x^2-2x+9=0\)

\(\Leftrightarrow x^2+2x-9=0\)

\(\Leftrightarrow x^2+2x+1=10\)

\(\Leftrightarrow\left(x+1\right)^2=10\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=\sqrt{10}\\x+1=-\sqrt{10}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{10}-1\\x=-\sqrt{10}-1\end{matrix}\right.\)

Vậy: \(S=\left\{\sqrt{10}-1;-\sqrt{10}-1\right\}\)

2) Ta có: \(5\left(2x-1\right)+7=4\left(2-x\right)+2\)

\(\Leftrightarrow10x-5+7=8-4x+2\)

\(\Leftrightarrow10x+4x=8+2+5-7\)

\(\Leftrightarrow14x=8\)

\(\Leftrightarrow x=\dfrac{4}{7}\)

Vậy: \(S=\left\{\dfrac{4}{7}\right\}\)

3) Ta có: \(x^2-6x+4\left(x-6\right)=0\)

\(\Leftrightarrow x\left(x-6\right)+4\left(x-6\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\)

Vậy: S={6;-4}

1: \(\Leftrightarrow2x^2-10x-3x-2x^2=0\)

=>-13x=0

=>x=0

2: \(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

=>3x=13

=>x=13/3

3: \(\Leftrightarrow4x^4-6x^3-4x^3+6x^3-2x^2=0\)

=>-2x^2=0

=>x=0

4: \(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

=>-8x=6-14=-8

=>x=1

`1)2x(x-5)-(3x+2x^2)=0`

`<=>2x^2-10x-3x-2x^2=0`

`<=>-13x=0`

`<=>x=0`

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`2)x(5-2x)+2x(x-1)=13`

`<=>5x-2x^2+2x^2-2x=13`

`<=>3x=13<=>x=13/3`

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`3)2x^3(2x-3)-x^2(4x^2-6x+2)=0`

`<=>4x^4-6x^3-4x^4+6x^3-2x^2=0`

`<=>x=0`

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`4)5x(x-1)-(x+2)(5x-7)=0`

`<=>5x^2-5x-5x^2+7x-10x+14=0`

`<=>-8x=-14`

`<=>x=7/4`

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`5)6x^2-(2x-3)(3x+2)=1`

`<=>6x^2-6x^2-4x+9x+6=1`

`<=>5x=-5<=>x=-1`

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`6)2x(1-x)+5=9-2x^2`

`<=>2x-2x^2+5=9-2x^2`

`<=>2x=4<=>x=2`

bài tập tết nâng cao phải ko

mk cũng có nhưng chưa làm dc

Bài 3:

a) \(\left(x-6\right).\left(2x-5\right).\left(3x+9\right)=0\)

\(\Leftrightarrow\left(x-6\right).\left(2x-5\right).3.\left(x+3\right)=0\)

Vì \(3\ne0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\2x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\2x=5\\x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\frac{5}{2}\\x=-3\end{matrix}\right.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{6;\frac{5}{2};-3\right\}.\)

b) \(2x.\left(x-3\right)+5.\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right).\left(2x+5\right)=0\)

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

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{3;-\frac{5}{2}\right\}.\)

c) \(\left(x^2-4\right)-\left(x-2\right).\left(3-2x\right)=0\)

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

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

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

\(\Leftrightarrow\left(x-2\right).\left(3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{3}\end{matrix}\right.\)

Vậy phương trình có tập hợp nghiệm là: \(S=\left\{2;\frac{1}{3}\right\}.\)

Chúc bạn học tốt!

Bài 4 xem lại đề nhé bác

a) \(\left(2x-1\right)^4+\left(2x-3\right)^4=0\)

\(\Leftrightarrow\left(2x-1\right)^4+\left(2x-3\right)^4=0^4\)

\(\Leftrightarrow\left(2x-1\right)+\left(2x-3\right)=0\)

\(\Rightarrow\hept{\begin{cases}2x-1=0\\2x-3=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=\left(0+1\right):2\\x=\left(0+3\right):2\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\x=\frac{3}{2}\end{cases}}\)

b) \(a^4-4a^3+12^2-16a+8=0\)

a)\(\left(2x-1\right)^4+\left(2x-3\right)^4=0\\ \)

do \(\left(2x-1\right)^4\ge0\)và \(\left(2x-3\right)^4\ge0\)

=> \(\hept{\begin{cases}\left(2x-1\right)^4=0\\\left(2x-3\right)^4=0\end{cases}\Rightarrow\hept{\begin{cases}2x-1=0\\2x-3=0\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{1}{2}\\x=\frac{3}{2}\end{cases}}}\).

Vậy x vừa = 1/2 vừa =3/2 => vô lý => không tồn tại nghiệm

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

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

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

\(\Leftrightarrow7x=1\)

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

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

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

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

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

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

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

`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):}`

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

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

a. \(9\left(x+2\right)-3\left(x+2\right)=0\)

\(\Leftrightarrow9x+18-3x-6=0\)

\(\Leftrightarrow6x+12=0\)

\(\Leftrightarrow x=-2\)

e. \(\left(2x-1\right)^2-45=0\)

\(\Leftrightarrow4x^2-2x+1-45=0\)

\(\Leftrightarrow4x^2-2x-44=0\)

Đến đó tự giải tiếp nha!

c. \(2\left(2x-5\right)-3x=0\)

\(\Leftrightarrow4x-10-3x=0\)

\(\Leftrightarrow x-10=0\)

\(\Leftrightarrow x=10\)

g. \(2x^2-6x=0\)

\(\Leftrightarrow2x\left(x-3\right)=0\)

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

b. Câu b bạn viết đề lại nhá!

d.\(\left(x-1\right)^2+x\left(5-x\right)=0\)

\(\Leftrightarrow x^2-2x+1+5x-x^2=0\)

\(\Leftrightarrow3x+1=0\)

\(\Leftrightarrow x=\dfrac{-1}{3}\)

e.

\(\left(2x-3\right)\left(2x+3\right)-\left(x+5\right)^2-3\left(x-1\right)\left(x+2\right)=0\)

\(\Leftrightarrow4x^2-9-x^2-10x-25-3x^2-3x+6=0\)

\(\Leftrightarrow-13x-28=0\)

\(\Leftrightarrow x=\dfrac{-28}{13}\)

a. \(x^2-7x+6=0\)

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

\(\Leftrightarrow\left(x^2-6x\right)-\left(x-6\right)=0\)

\(\Leftrightarrow x\left(x-6\right)-\left(x-6\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=6\end{matrix}\right.\)