a: \(\Leftrightarrow x^2-7x^2+28x=16\)

\(\Leftrightarrow-6x^2+28x-16=0\)

\(\Leftrightarrow3x^2-14x+8=0\)

\(\text{Δ}=\left(-14\right)^2-4\cdot3\cdot8=100\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{14-10}{6}=\dfrac{4}{6}=\dfrac{2}{3}\\x_2=\dfrac{14+10}{6}=\dfrac{24}{6}=4\end{matrix}\right.\)

a) x²-7x(x-4)=16

 x²-16-7x(x-4)=0

(x-4)(x+4)-7x(x-4)=0

(x-4)(x+4-7x)=0

(x-4)(-6x+4)=0

=>x=4 hoặc x=2/3

b) (x+5)^2-3(x+1)=22

x²+10x+25-3x-3=22

x²+10x-3x=22+3-25

x²+7x=0

x(x+7)=0

=>x=0 hoặc x=-7

f: Ta có: \(16x^2-9\left(x+1\right)^2=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\)

3/2 - 4(1/4 -x) = 2/3 -7x

3/2 - 1 + 4x = 2/3 -7x

4x + 7x = 2/3 - 3/2 + 1

11x = 1/6

x = 66

 x=1 nha bạn

a. x=1

b. x=1

a. x = 1

b. x = 1

a,x=\(\frac{4}{3}\)

b, x= 1

c, x= viết đề rõ hơn

d, \(\frac{13}{6}\)

Giải : 

\(\frac{x+1}{x-2}=\frac{3}{4}\)

\(\Rightarrow4.\left(x-1\right)=3.\left(x-2\right)\)

\(\Rightarrow4x-4=3x-6\)

\(\Rightarrow4x-4-3x+6=0\)

\(\Rightarrow x+2=0\)

\(\Rightarrow x=-2\)Không thỏa mãn => Không có giá trị x thỏa mãn đề bài 

\(\frac{2x-3}{x+1}=\frac{4}{7}\)

\(\Rightarrow7.\left(2x-3\right)=4.\left(x+1\right)\)

\(\Rightarrow14x-21-4x-4=0\)

\(\Rightarrow10x-25=0\)

\(\Rightarrow10x=25\)

\(\Rightarrow x=\frac{25}{10}=\frac{5}{2}\)

Giá trị trên thỏa mãn đầu bài

Các phần khác em làm tương tự nha

\(a,=x^2-4x+4-\dfrac{15}{4}=\left(x-2\right)^2-\dfrac{15}{4}=\left(x-2-\dfrac{\sqrt{15}}{2}\right)\left(x-2+\dfrac{\sqrt{15}}{2}\right)\\ b,=?\\ c,\Rightarrow x^2+7x-8=0\\ \Rightarrow\left(x+8\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=-8\\x=1\end{matrix}\right.\\ d,Sửa:x^3-3x^2=-27+9x\\ \Rightarrow x^3-3x^2+9x-27=0\\ \Rightarrow x^2\left(x-3\right)+9\left(x-3\right)=0\\ \Rightarrow\left(x^2+9\right)\left(x-3\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-9\left(vô.lí\right)\\x=3\end{matrix}\right.\\ \Rightarrow x=3\\ e,\Rightarrow x\left(x-3\right)-7x+21=0\\ \Rightarrow x\left(x-3\right)-7\left(x-3\right)=0\\ \Rightarrow\left(x-7\right)\left(x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\\ f,\Rightarrow x^2\left(x-2\right)+\left(x-2\right)=0\\ \Rightarrow\left(x^2+1\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=2\end{matrix}\right.\\ \Rightarrow x=2\)

\(g,\Rightarrow x^2-4x+4=0\\ \Rightarrow\left(x-2\right)^2=0\\ \Rightarrow x=2\\ h,Sửa:x^3-x^2+x=1\\ \Rightarrow x^2\left(x-1\right)+\left(x-1\right)=0\\ \Rightarrow\left(x^2+1\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=1\end{matrix}\right.\\ \Rightarrow x=1\)

a: \(\Leftrightarrow\left|x-1\right|=3-2x\)

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

=>x=-2/3

b: Trường hợp 1: x<-3

Pt sẽ là:

\(-x-1-x-3=10-4x\)

=>-2x-4=10-4x

=>2x=14

hay x=7(loại)

Trường hợp 2: -3<=x<-1

Pt sẽ là \(x+3-x-1=10-4x\)

=>10-4x=2

=>4x=8

hay x=2(loại)

Trường hợp 3: x>=-1

Pt sẽ là x+1+x+3=10-4x

=>2x+4=10-4x

=>6x=6

hay x=1(nhận)