Thiếu rồi

\(\left(5x-2\right)\left(2x+7\right)-4x^2-25=0\)

\(10x+35-4x^2-14x-4x^2+25=0\)

\(-4x+60-8x^2=0\)

\(-4\left(2x^2+x-15\right)=0\)

\(-4\left(2x^2+6x-5x-15\right)=0\)

\(-4\left(2x-5\right)\left(x+3\right)=0\)

=> \(x\) ∈ \(\left\{\dfrac{5}{2};-3\right\}\)

a) \(4x^2-12x=-9\)

\(\Leftrightarrow4x^2-12x+9=0\)

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

\(\Leftrightarrow2x-3=0\Leftrightarrow x=\frac{3}{2}\)

b) \(\left(5-2x\right)\left(2x+7\right)=4x^2-25\)

\(\Leftrightarrow\left(5-2x\right)\left(2x+7\right)+\left(25-4x^2\right)=0\)

\(\Leftrightarrow\left(5-2x\right)\left(2x+7\right)+\left(5-2x\right)\left(5+2x\right)=0\)

\(\Leftrightarrow\left(5-2x\right)\left(2x+7+5+2x\right)=0\)

\(\Leftrightarrow\left(5-2x\right)\left(4x+12\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-3\end{array}\right.\)

c)\(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)

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

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

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

d) \(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)

\(\Leftrightarrow\left[2\left(2x+7\right)-3\left(x+3\right)\right]\left[2\left(2x+7\right)+3\left(x+3\right)\right]=0\)

\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-5\\x=-\frac{23}{17}\end{array}\right.\)

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`

Ta có:

\(a)\left(2x-5\right)\left(x+2\right)-2x\left(x-1\right)=15\)

\(\Leftrightarrow\left(2x^2-x-10\right)-\left(2x^2-2x\right)=15\Leftrightarrow x-10=15\)

\(\Leftrightarrow x=25\)

\(b)\left(5-2x\right)\left(2x+7\right)=4x^2-25\)

\(\Leftrightarrow\left(5-2x\right)\left(2x+7\right)=\left(2x-5\right)\left(2x+5\right)\)

\(\Leftrightarrow\left(5-2x\right)\left(4x+12\right)=0\)

\(5-2x=0\Leftrightarrow x=\frac{5}{2}\)

\(4x+12=0\Leftrightarrow x=-3\)

Vậy ..........................................

a/\(x^3-9x=0\Leftrightarrow x\left(x^2-9\right)=0\Leftrightarrow x\left(x-3\right)\left(x+3\right)=0\)

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

b/ \(\left(5-2x\right)\left(2x+7\right)=4x^2-25\Leftrightarrow\left(5-2x\right)\left(2x+7\right)=\left(2x-5\right)\left(2x+5\right)\)

\(\Leftrightarrow\left(2x-5\right)\left(2x+5+2x+7\right)=0\Leftrightarrow\left(2x-5\right)\left(4x+12\right)=0\) 

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-3\end{array}\right.\)

\(x^3-9x=0\)

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

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

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

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

b ) \(\left(5-2x\right)\left(2x+7\right)=4x^2-25\)

\(\Leftrightarrow-\left(2x-5\right)\left(2x+7\right)-\left(2x-5\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(-2x-7-2x-5\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(-4x-12\right)=0\)

\(\Leftrightarrow-4\left(2x-5\right)\left(x+3\right)=0\)

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

x^3 - 9x = 0

x(x^2 - 9) = 0

x = 0 hoặc x^2 - 9 = 0

x = 0 hoặc x^2 - 3^2 = 0

x = 0 hoặc x - 3 = 0 hoặc x + 3 = 0

vậy x = 0 hoặc x = 3 hoặc x= -3

\(B1\\ a,2x+10y=2\left(x+5y\right)\\ b,x^2+4x+4=x^2+2.2x+2^2=\left(x+2\right)^2\\ c,x^2-y^2+10y-25\\ =\left(x^2-y^2\right)+5\left(2y-5\right)\\ =\left(x-y\right)\left(x+y\right)+5\left(2y-5\right)\\ B2\)

\(a,x^2-3x+x-3=0\\ =>x\left(x-3\right)+\left(x-3\right)=0\\ =>\left(x+1\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x+1=0\\x-3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\\ b,2x\left(x-3\right)-\dfrac{1}{2}\left(4x^2-3\right)=0\\ =>2x^2-6x-2x^2+\dfrac{3}{2}=0\\ =>-6x=-\dfrac{3}{2}\\ =>x=\left(-\dfrac{3}{2}\right):\left(-6\right)\\ =>x=\dfrac{1}{4}\\ c,x^2-\left(x-3\right)\left(2x-5\right)=9\\ =>x^2-2x^2+6x+5x-15=9\\ =>-x^2+11-15-9=0\\ =>-x^2+11x-24=0\\ =>-x^2+8x+3x-24=0\\ =>-x\left(x-8\right)+3\left(x-8\right)=0\\ =>\left(3-x\right)\left(x-8\right)=0\\ =>\left[{}\begin{matrix}3-x=0\\x-8=0\end{matrix}\right.=>\left[{}\begin{matrix}x=3\\x=8\end{matrix}\right.\)

bài tập tết nâng cao phải ko

mk cũng có nhưng chưa làm dc

Sửa đề: \(4x^2-25+\left(2x-5\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x+5\right)+\left(2x-5\right)\left(2x+7\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\cdot\left(4x+12\right)=0\)

=>x=5/2 hoặc x=-3

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

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

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

b) \(\left(2x-1\right)\left(3x+2\right)\left(4x-5\right)\left(x-7\right)=0\)

\(\Leftrightarrow\left[\begin{matrix}2x-1=0\\3x+2=0\\4x-5=0\\x-7=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=\frac{1}{2}\\x=-\frac{2}{3}\\x=\frac{5}{4}\\x=7\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là \(S=\left\{\frac{1}{2};-\frac{2}{3};\frac{5}{4};7\right\}\)

c) \(x^2-6x+11=0\)

\(\Leftrightarrow x^2-6x+9+2=0\)

\(\Leftrightarrow\left(x-3\right)^2+2=0\) (vô lí)

Vậy phương trình vô nghiệm

d) \(\left(x^2+2x+3\right)\left(x^2-25\right)\left(x+19\right)=0\)

\(\Leftrightarrow\left(x^2+2x+1+2\right)\left(x+5\right)\left(x-5\right)\left(x+19\right)=0\)

\(\Leftrightarrow\left[\left(x+1\right)^2+2\right]\left(x+5\right)\left(x-5\right)\left(x+19\right)=0\)

\(\Leftrightarrow\left[\begin{matrix}x+5=0\\x-5=0\\x+19=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=-5\\x=5\\x=-19\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là \(S=\left\{\pm5;-19\right\}\)

a,b,d dễ mà bạn tự làm

c,x²-6x+11=0<=> x²-6x+9+2=0

<=>(x-3)²=-2(vô lý)

vậy pt vô nghiệm