a) \(22-x\left(1-4x\right)=\left(2x+3\right)^3\)

\(\Leftrightarrow22-x+4x^2=8x^3+36x^2+54x+27\)

\(\Leftrightarrow-x-54x+4x^2-36x^2-8x^3=-22+27\)

\(\Leftrightarrow-8x^3-32x^2-55x=5\Leftrightarrow-8x^3-32x^2-55x-5=0\)

Bn tự làm tiếp nhé

b) \(\frac{2x}{3}+\frac{2x-1}{6}=\frac{4-x}{3}\Leftrightarrow\frac{2.2x}{6}+\frac{2x-1}{6}=\frac{2\left(4-x\right)}{6}\)

\(\Leftrightarrow2.2x+2x-1=2\left(4-x\right)\Leftrightarrow4x+2x-1=8-2x\)

\(\Leftrightarrow6x-1=8-2x\Leftrightarrow8x=9\Leftrightarrow x=\frac{9}{8}\)

Vậy phương trình có tập nghiệm S ={9/8}

c) \(\frac{x-1}{2019}+\frac{x-2}{2018}=\frac{x-3}{2017}+\frac{x-4}{2016}\)

\(\Leftrightarrow\left(\frac{x-1}{2019}-1\right)+\left(\frac{x-2}{2018}-1\right)=\left(\frac{x-3}{2017}-1\right)+\left(\frac{x-4}{2016}-1\right)\)

\(\Leftrightarrow\frac{x-2020}{2019}+\frac{x-2020}{2018}-\frac{x-2020}{2017}-\frac{x-2020}{2016}=0\)

\(\Leftrightarrow\left(x-2020\right)\left(\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}\right)=0\)

Do \(\frac{1}{2019}+\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}>0\)

Nên \(x-2020=0\Leftrightarrow x=2020\)

Không chắc đúng hay không nha,tui mới lớp 7=(

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

\(\Leftrightarrow\left(a+b\right)\left(a-b\right)x+b\left(a+b\right)=0\)

\(\Leftrightarrow\left(a+b\right)\left[\left(a-b\right)x+b\right]=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=-b\\ax-bx+b=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}a=-b\\x=-\frac{b}{a-b}\end{cases}}\)

+Với a = -b,thì phương trình trở thành:

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

\(\Leftrightarrow0=0\) (luôn đúng)

Vậy nếu a = -b thì phương trình có vô số nghiệm.

Với ax - bx + b = 0 thì \(x=-\frac{b}{a-b}=\frac{b}{b-a}\)

Ta có \(a+b+c=1\)

\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ca=1\)

Mà \(a^2+b^2+c^2=1\)

\(\Rightarrow ab+bc+ca=0\)

Đặt \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=k\)

\(\Rightarrow\hept{\begin{cases}x=ak\\y=bk\\z=ck\end{cases}}\)

\(\Rightarrow xy+yz+zx=abk^2+bck^2+cak^2\)

\(\Rightarrow xy+yz+zx=k^2\left(ab+bc+ca\right)\)

Mà \(ab+bc+ca=0\)(theo trên)

\(\Rightarrow xy+yz+zx=0\left(đpcm\right)\)

\(x- \frac {\frac x 2-3+\frac x 4} 2=\frac {2x-\frac {10-7x} 3} 2-(x-1)\)

<=> \(\frac {2x- \frac {10-7x} {3} + \frac {x} {2} -3 + \frac {x} {4}} 2 - 2x+1=0\)

<=>\(2x-\frac{10-7x} 3+\frac x 2-3+\frac x 4-4x+2=0\)

<=>\(24x-40+28x+6x-36+3x-48x+24=0\)

<=>\(13x-52=0 <=>x=4 \)

S={4}

Đặt \(\frac{2a}{b+c}=t\)(t>0).Ta cần c/m: \(t+\frac{1}{t}\ge2\) (t > 0)

Thật vậy,áp dụng BĐT AM-GM (Cô si),ta có \(t+\frac{1}{t}\ge2\sqrt{t.\frac{1}{t}}=2\)

Dấu "=" xảy ra \(\Leftrightarrow t^2=1\Leftrightarrow t=1\Leftrightarrow2a=b+c\Leftrightarrow a=b=c\)

\(-x^5+4x^4=-12x^3\)

\(\Leftrightarrow x^5-4x^4-12x^3=0\)

\(\Leftrightarrow x^3\left(x^2-4x-12\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-4x-12=0\left(1\right)\end{cases}}\)

\(Pt\left(1\right)\Leftrightarrow\left(x+2\right)\left(x-6\right)=0\)

              \(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=6\end{cases}}\)

Vậy \(x\in\left\{-2;0;6\right\}\)

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