a: =(x-3)(2x+5)

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

=>(x-2)(5-x)=0

=>x=2 hoặc x=5

c: =>x-1=0

hay x=1

TK

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

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

dùng hđt \(a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)\)

có nhân tử chung

Đặt \(x^2+3x-4=a;2x^2-5x+3=b\)

Ta có phương trình: \(a^3+b^3=\left(a+b\right)^3\)

=>3ab(a+b)=0

\(\Leftrightarrow\left(x^2+3x-4\right)\left(2x^2-5x+3\right)\left(3x^2-2x-1\right)=0\)

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

hay \(x\in\left\{-4;1;\dfrac{3}{2};-\dfrac{1}{3}\right\}\)

Lời giải :

Đặt \(\hept{\begin{cases}x^2+3x-4=a\\2x^2-5x+3=b\end{cases}}\)

\(\Rightarrow a+b=\left(x^2+3x-4\right)+\left(2x^2-5x+3\right)=3x^2-2x-1\)

Khi đó phương trình đã cho trở thành :

\(a^3+b^3=\left(a+b\right)^3\)

\(\Leftrightarrow a^3+b^3=a^3+b^3+3ab.\left(a+b\right)\)

\(\Leftrightarrow3ab.\left(a+b\right)=0\) \(\Rightarrow\orbr{\begin{cases}a+b=0\\ab=0\end{cases}}\)

+) Với \(a+b=0\Rightarrow3x^2-2x-1=0\)

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

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

+) Với \(ab=0\Rightarrow\left(x^2+3x-4\right).\left(2x^2-5x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+3x-4=0\left(1\right)\\2x^2-5x+3=0\left(2\right)\end{cases}}\)

Pt (1) \(\Leftrightarrow\left(x-1\right)\left(x+4\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-4\end{cases}}\)

Pt (2) \(\Leftrightarrow\left(x-1\right)\left(2x-3\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{3}{2}\end{cases}}\)

Vạy phương trình đã cho có tập nghiệm \(S=\left\{-4,-\frac{1}{3},1,\frac{3}{2}\right\}\)

bậc nhất môt ẩn đây ak

Bài 1:

a) Ta có: \(\frac{4}{5}x-3=\frac{1}{5}x\left(4x-15\right)\)

\(\Leftrightarrow\frac{4x}{5}-3=\frac{4x^2}{5}-3x\)

\(\Leftrightarrow\frac{12x}{15}-\frac{45}{15}-\frac{12x^2}{15}+\frac{45x}{15}=0\)

Suy ra: \(12x-45-12x^2+45x=0\)

\(\Leftrightarrow-12x^2+57x-45=0\)

\(\Leftrightarrow-12x^2+12x+45x-45=0\)

\(\Leftrightarrow-12x\left(x-1\right)+45\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(-12x+45\right)=0\)

\(\Leftrightarrow-3\left(x-1\right)\left(4x-15\right)=0\)

mà \(-3\ne0\)

nên \(\left[{}\begin{matrix}x-1=0\\4x-15=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\4x=15\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{15}{4}\end{matrix}\right.\)

Vậy: Tập nghiệm \(S=\left\{1;\frac{15}{4}\right\}\)

b) Ta có: \(\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}=\frac{\left(x-3\right)\left(3-x\right)}{4}\)

\(\Leftrightarrow\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}+\frac{\left(x-3\right)^2}{4}=0\)

\(\Leftrightarrow\frac{12\left(x-3\right)}{12}-\frac{2\left(x-3\right)\left(2x-5\right)}{12}+\frac{3\left(x-3\right)^2}{12}=0\)

Suy ra: \(12\left(x-3\right)-2\left(2x^2-11x+15\right)+3\left(x^2-6x+9\right)=0\)

\(\Leftrightarrow12x-36-4x^2+22x-30+3x^2-18x+27=0\)

\(\Leftrightarrow-x^2+16x-39=0\)

\(\Leftrightarrow-\left(x^2-16x+39\right)=0\)

\(\Leftrightarrow x^2-13x-3x+39=0\)

\(\Leftrightarrow x\left(x-13\right)-3\left(x-13\right)=0\)

\(\Leftrightarrow\left(x-13\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-13=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=13\\x=3\end{matrix}\right.\)

Vậy: Tập nghiệm S={3;13}

c) Ta có: \(\frac{\left(3x+1\right)\left(3x-2\right)}{3}+5\left(3x+1\right)=\frac{2\left(2x+1\right)\left(3x+1\right)}{3}+2x\left(3x+1\right)\)

\(\Leftrightarrow\frac{9x^2-3x-2}{3}+5\left(3x+1\right)-\frac{12x^2+10x+2}{3}-2x\left(3x+1\right)=0\)

\(\Leftrightarrow\frac{9x^2-3x-2-12x^2-10x-2}{3}-6x^2+13x+5=0\)

\(\Leftrightarrow\frac{-3x^2-13x-4}{3}+\frac{3\left(-6x^2+13x+5\right)}{3}=0\)

Suy ra: \(-3x^2-13x-4-18x^2+39x+15=0\)

\(\Leftrightarrow-21x^2+26x+11=0\)

\(\Leftrightarrow-21x^2-7x+33x+11=0\)

\(\Leftrightarrow-7x\left(3x+1\right)+11\left(3x+1\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(-7x+11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\-7x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\-7x=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{3}\\x=\frac{11}{7}\end{matrix}\right.\)

Vậy: Tập nghiệm \(S=\left\{-\frac{1}{3};\frac{11}{7}\right\}\)