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

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

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

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

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

\(\frac{2}{2x+3}+\frac{5}{2x-3}-\frac{2x-33}{9-4x^2}\)

= \(\frac{2}{2x+3}+\frac{5}{2x-3}+\frac{2x-33}{4x^2-9}\)

= \(\frac{2\left(2x-3\right)}{\left(2x+3\right)\left(2x-3\right)}+\frac{5\left(2x+3\right)}{\left(2x-3\right)\left(2x+3\right)}+\frac{2x-33}{\left(2x-3\right)\left(2x+3\right)}\)

= \(\frac{4x-6+10x-15+2x-33}{\left(2x-3\right)\left(2x+3\right)}\)

= \(\frac{16x-54}{\left(2x-3\right)\left(2x+3\right)}\)

\(\frac{2}{2x+3}+\frac{5}{2x-3}-\frac{2x-33}{9-4x^2}\)\(=\frac{2}{2x+3}+\frac{5}{2x-3}+\frac{2x-33}{4x^2-9}\)

\(=\frac{2\left(2x-3\right)}{\left(2x+3\right)\left(2x-3\right)}+\frac{5\left(2x+3\right)}{\left(2x+3\right)\left(2x-3\right)}+\frac{2x-33}{\left(2x+3\right)\left(2x-3\right)}\)

\(=\frac{4x-6+10x+15+2x-33}{\left(2x+3\right)\left(2x-3\right)}=\frac{16x-24}{\left(2x+3\right)\left(2x-3\right)}=\frac{8\left(2x-3\right)}{\left(2x+3\right)\left(2x-3\right)}=\frac{8}{2x+3}\)

giải bất phương trình

a: =>-4x>16

=>x<-4

c: =>20x-25<=21-3x

=>23x<=46

=>x<=2

d: =>20(2x-5)-30(3x-1)<12(3-x)-15(2x-1)

=>40x-100-90x+30<36-12x-30x+15

=>-50x-70<-42x+51

=>-8x<121

=>x>-121/8

\(\frac{4}{2x+3}-\frac{7}{3x-5}=0\left(đkxđ:x\ne-\frac{3}{2};\frac{5}{3}\right)\)

\(< =>\frac{4\left(3x-5\right)}{\left(2x+3\right)\left(3x-5\right)}-\frac{7\left(2x+3\right)}{\left(2x+3\right)\left(3x-5\right)}=0\)

\(< =>12x-20-14x-21=0\)

\(< =>2x+41=0< =>x=-\frac{41}{2}\left(tm\right)\)

\(\frac{4}{2x-3}+\frac{4x}{4x^2-9}=\frac{1}{2x+3}\left(đk:x\ne-\frac{3}{2};\frac{3}{2}\right)\)

\(< =>\frac{4\left(2x+3\right)}{\left(2x-3\right)\left(2x+3\right)}+\frac{4x}{\left(2x-3\right)\left(2x+3\right)}-\frac{2x-3}{\left(2x+3\right)\left(2x-3\right)}=0\)

\(< =>8x+12+4x-2x+3=0\)

\(< =>10x=15< =>x=\frac{15}{10}=\frac{3}{2}\left(ktm\right)\)

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

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

\(\Leftrightarrow4x^2-4x^2+14x-10x=35-25\)

\(\Leftrightarrow4x=10\)

\(\Leftrightarrow x=\dfrac{10}{4}=\dfrac{5}{2}\)

Vậy \(x=\dfrac{5}{2}\)

2) \(x^2-4x+5\)

\(=-(4x-x^2-5 )\)

\(= -[-(x^2-4x)-5 ]\)

\(=-[ -(x^2-2x.2+4-4)-5 ]\)

\(= -[-(x-2)^2+4-5 ]\)

\(= -[-(x-2)^2-1 ]\)

Vì \(-(x-2)^2 ≤0\)\(\forall x\) \(\Rightarrow\) \(-(x-2)^2-1<0\) \(\forall x\)

\(\Rightarrow\)\(-[-(x-2)^2-1 ]>0\)\(\forall x\)

\(\Rightarrow x^2-4x+5>0\)\(\forall x\)

2

\(x^2-4x+5=x^2-4x+4+1\\ =\left(x-2\right)^2+1>0\)

nếu sai ở đâu sửa giúp mình với

nhanh mình đang cần gấp

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

\(\Rightarrow\left(2x\right)^2-5^2=\left(2x-5\right).\left(2x+7\right)\)

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

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

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

Vì \(-2\ne0.\)

\(\Rightarrow2x-5=0\)

\(\Rightarrow2x=0+5\)

\(\Rightarrow2x=5\)

\(\Rightarrow x=5:2\)

\(\Rightarrow x=\frac{5}{2}\)

Vậy \(x=\frac{5}{2}.\)

Chúc bạn học tốt!

4x²−25 - ( 4x² + 14x -10x - 35) = 0

4x² −25 - 4x² -14x + 10x + 35 = 0

-4x + 10 = 0

=> x = -\(\frac{5}{2}\)

Vậy x = -\(\frac{5}{2}\)

Câu này bình thường mà bạn có thẻ giải bằng cách dùng hàng đẳng thức số 3 hoặc nhân tung vào như vậy cũng đc

       x²-4x+4=4x²-12x+9

\(\Leftrightarrow\)3x²-8x+5=0

\(\Leftrightarrow\)3x²-3x-5x+5=0

\(\Leftrightarrow\)3x(x-1)-5(x-1)=0

\(\Leftrightarrow\)(x-1)(3x-5)=0

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{5}{3}\\x=1\end{cases}}\)

b,x²-2x-25=0

\(\Leftrightarrow\)(x-1)²-26=0

\(\Leftrightarrow\)(x-1-\(\sqrt{26}\))(x-1+\(\sqrt{26}\))=0

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\sqrt{26}+1\\x=-\sqrt{26}+1\end{cases}}\)

2, a, x^2-2x+1+4=(x-1)^2+4\(\ge\)4

b, 4x^2-4x+1-1+y^2+2y+1-1-2015=(2x-1)^2+(y+1)^2-2017\(\ge\)-2017

mk làm như thế thôi chứ bài kia dài quá mk làm biếng sory

Nguyễn Thị Hà Tiên : Cảm ơn bạn nhiều lắm =)) Mik đã bt hướng làm bài rồi :3 Thực sự cảm ơn pạn nek <3