a)2x.(3x+5)-x.(6x-1)=33

=>\(6x^2+10x-6x^2+x=33\)

=>11x=33

=>x=3

b)x(3x-1)+12x-4=0

=>x(3x-1)+4(3x-1)=0

=>(x-4)(3x-1)=0

=>x-4=0 hoặc 3x-1=0

+)x-4=0 +)3x-1=0

=>x=4 =>x=\(\frac{1}{3}\)

c)5x(2x+1)-12x-6=0

=>10x\(^2\)+5x-12x-6=0

=>10x\(^2\)-7x-6=0

=>(10x\(^2\)+5x)-(12x+6)=0

=>5x(2x+1)-6(2x+1)=0

=>(5x-6)(2x+1)=0

=>\(\left[{}\begin{matrix}5x-6=0\\2x+1=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=\frac{6}{5}\\x=\frac{-1}{2}\end{matrix}\right.\)

a) Ta có: \(x^2-9x+20=0\)

\(\Leftrightarrow x^2-5x-4x+20=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\end{matrix}\right.\)

Vậy: x∈{4;5}

b) Ta có: \(x^3-4x^2+5x=0\)

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

Ta có: \(x^2-4x+5\)

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

Ta có: \(\left(x-2\right)^2\ge0\forall x\)

\(\Rightarrow\left(x-2\right)^2+1\ge1>0\forall x\)

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

Từ (1) và (2) suy ra x=0

Vậy: x=0

c) Sửa đề: \(x^2-2x-15=0\)

Ta có: \(x^2-2x-15=0\)

\(\Leftrightarrow x^2+3x-5x-15=0\)

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

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

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

Vậy: x∈{-3;5}

d) Ta có: \(\left(x^2-1\right)^2=4x+1\)

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

\(\Leftrightarrow x^4-2x^2-4x=0\)

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

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

\(\Leftrightarrow x\cdot\left[x\left(x^2+2x+2\right)-2\left(x^2+2x+2\right)\right]=0\)

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

Ta có: \(x^2+2x+2\)

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

Ta có: \(\left(x+1\right)^2\ge0\forall x\)

\(\Rightarrow\left(x+1\right)^2+1\ge1>0\forall x\)

hay \(x^2+2x+2>0\forall x\)(4)

Từ (3) và (4) suy ra

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

Vậy: x∈{0;2}

a) \(5x\left(x+4\right)-x\left(5x+1\right)=0\)

\(\Leftrightarrow x\left[5\left(x+4\right)-5x-1\right]=0\)

\(\Leftrightarrow x\left(5x+20-5x-1\right)=0\Leftrightarrow x=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=\frac{4}{3}\end{cases}}\)

c) \(x\left(x-3\right)+4x-12=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-4\end{cases}}\)

d) \(x^2-36=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)

e) \(x^2+3x+1=2\)

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

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

\(\Leftrightarrow x^2+3x+\frac{3}{2}-\frac{5}{2}=0\)

\(\Leftrightarrow\left(x+\frac{3}{2}\right)^2-\frac{5}{2}=0\)

\(\Leftrightarrow\left(x+\frac{3}{2}+\frac{\sqrt{5}}{\sqrt{2}}\right)\left(x+\frac{3}{2}-\frac{\sqrt{5}}{\sqrt{2}}\right)=0\)

Còn lại ........... Tự lm nất nha 

Câu 1:

a) 1 + x = 0

c) 1 - 2t = 0

d) 3y = 0

câu 1 bạn tự làm nha

câu 2:

a) 4x-20=0

<=> 4x =20

<=>x =\(\dfrac{20}{4}\)

<=> x=5

b) 2x+x+12=0

<=>3x+12=0

<=>3x =-12

<=>x =\(\dfrac{-12}{3}\)

<=> x=-4

c) x-5=3-x

<=>x+x=3+5

<=>2x=8

<=>x =\(\dfrac{8}{2}\)

<=>x=4

d) 7-3x=9-x

<=>-3x+x =9-7

<=>-2x=2

<=>x=\(\dfrac{2}{-2}\)

<=>x=-1

\(a,\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)

\(\left(x-1\right)\left(5x+3-3x+8\right)=0\)

\(\left(x-1\right)\left(2x+11\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\2x+11=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\2x=-11\end{cases}\Rightarrow}\orbr{\begin{cases}x=1\\x=-\frac{11}{2}\end{cases}}}\)

\(b,3x\left(25x+15\right)-35\left(5x+3\right)=0\)

\(15x\left(5x+3\right)-35\left(5x+3\right)=0\)

\(\left(5x+3\right).5\left(3x-7\right)=0\)

\(\Rightarrow\orbr{\begin{cases}5x+3=0\\5\left(3x-7\right)=0\end{cases}\Rightarrow\orbr{\begin{cases}5x=-3\\3x-7=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{3}{5}\\3x=7\end{cases}\Rightarrow}\orbr{\begin{cases}x=-\frac{3}{5}\\x=\frac{7}{3}\end{cases}}}\)

\(c,\left(2-3x\right)\left(x+11\right)=\left(3x-2\right)\left(2-5x\right)\)

\(\left(3x-2\right)\left(2-5x\right)+\left(3x-2\right)\left(x+11\right)=0\)

\(\left(3x-2\right)\left(2-5x+x+11\right)=0\)

\(\left(3x-2\right)\left(13-4x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}3x-2=0\\13-4x=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=2\\4x=13\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{13}{4}\end{cases}}}\)

còn đâu tự lm lười :_# 

a) \(2.\left(x+5\right)-x^2-5x=0\)

\(\Leftrightarrow2.\left(x+5\right)-x.\left(x+5\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\2-x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}\)

Vậy \(S=\left\{-5,2\right\}\)

b) \(x^3-5x^2-4x+20=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\x^2-4=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=5\\x=\pm2\end{cases}}\)

Vậy \(S=\left\{5,\pm2\right\}\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\3x+2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=4\\x=-\frac{3}{2}\end{cases}}\)

Vậy \(S=\left\{4,-\frac{3}{2}\right\}\)

a) \(x\left(x-5\right)-4x+20=0\)

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

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-4=0\\x-5=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=4\\x=5\end{array}\right.\)

b) \(x\left(x+6\right)-7x-42=0\)

\(\Leftrightarrow x\left(x+6\right)-7\left(x+6\right)=0\)

\(\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x+6=0\\x-7=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-6\\x=7\end{array}\right.\)

d) \(x^2-9x+8=0\)

\(\Leftrightarrow x^2-x-8x+8=0\)

\(\Leftrightarrow x\left(x-1\right)-8\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-8\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x-8=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=8\end{array}\right.\)

g) \(3x^2-5x+2=0\)

\(\Leftrightarrow3x^2-3x-2x+2=0\)

\(\Leftrightarrow3x\left(x-1\right)-2\left(x-1\right)=0\)

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

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

\(a,2\left(x+5\right)-x^2-5x=0\)

\(< =>2x+10-x^2-5x=0\)

\(< =>-x^2-3x+10=0\)

\(< =>-\left(x^2+3x+\frac{9}{4}\right)+\frac{49}{4}=0\)

\(< =>-\left(x+\frac{3}{2}\right)^2=-\frac{49}{4}\)

\(< =>\left(x+\frac{3}{2}\right)^2=\frac{49}{4}< =>\orbr{\begin{cases}x+\frac{3}{2}=\sqrt{\frac{49}{4}}\\x+\frac{3}{2}=-\sqrt{\frac{49}{4}}\end{cases}}\)

\(< =>\orbr{\begin{cases}x=\frac{7}{2}-\frac{3}{2}=\frac{4}{2}=2\\x=-\frac{7}{2}-\frac{3}{2}=-\frac{10}{2}=-5\end{cases}}\)

b, Đật x = y+5/3 khi đó phương trình trở thành 

\(y^3-\frac{37}{3}y+\frac{476}{27}=0\)

Đặt \(y=u+v\)sao cho uv=37/9 thế vào ta được phương trình mới sau ta được

\(u^3+v^3+\left(3uv-\frac{37}{3}\right)\left(u+v\right)+\frac{426}{27}=0\)

Khi đó ta có hệ sau : \(\hept{\begin{cases}u^3+v^3=-\frac{426}{27}\\u^3v^3=\frac{50653}{729}\end{cases}}\)

Theo Vi ét u^3 và v^3 là 2 nghiệm của pt \(x^2-\frac{426}{27}x+\frac{50653}{729}=0\)

Đến đây delta phát rồi tìm ngược lại là xong :))))

mình dùng cardano nhưng làm trong nháp xong gửi nên chắc chắc bạn sẽ không hiểu được :V

làm luôn câu cuối nhé ^^

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

\(\Leftrightarrow\left(4x^2-4x+1\right)-\left(x^2+6x+9\right)=0\)

\(\Leftrightarrow4x^2-4x+1-x^2-6x-9=0\)

\(\Leftrightarrow3x^2-10x-8=0\)

\(\Leftrightarrow3\left(x^2-\frac{10}{3}x+\frac{25}{9}\right)-\frac{147}{9}=0\)

\(\Leftrightarrow3\left(x-\frac{5}{3}\right)^2=\frac{147}{9}\Leftrightarrow\left(x-\frac{5}{3}\right)^2=\frac{147}{27}\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{5}{3}=\sqrt{\frac{147}{27}}\\x-\frac{5}{3}=-\sqrt{\frac{147}{27}}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\sqrt{\frac{147}{27}}+\frac{5}{3}\\x=-\sqrt{\frac{147}{27}}+\frac{5}{3}\end{cases}}\)

bạn có thể giải rõ giúp mình phần b được không ?

\(A.\left(2,3x-6,5\right)\left(0,1x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2,3x-6,5=0\\0,1x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}2,3x=6,5\\0,1x=-2\end{cases}}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{6,5}{2,3}\\x=-20\end{cases}}\)