Bài giải

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

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

\(\frac{2}{7}x=-\frac{5}{4}\)

\(x=-\frac{5}{4}\text{ : }\frac{2}{7}\)

\(x=-\frac{35}{8}\)

b, \(\left(6x+\frac{2}{5}\right)=-\frac{8}{125}\)

\(6x=-\frac{8}{125}-\frac{2}{5}\)

\(6x=-\frac{58}{125}\)

\(x=-\frac{58}{125}\text{ : }6\)

\(x=\frac{-29}{375}\)

c, \(\left|x-\frac{2}{3}\right|\cdot\left(18-6x^2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\left|x-\frac{2}{3}\right|=0\\18-6x^2=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x-\frac{2}{3}=0\\6x^2=18\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x^2=3\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\sqrt{3}\end{cases}}\)

\(\Rightarrow\text{ }x\in\left\{\frac{2}{3}\text{ ; }\sqrt{3}\right\}\)

ai làm cho mik với pls

Ukm

It's very hard

l can't do it 

Sorry!

a) \(x^4-x^3-7x^2+x+6=0\)

\(\Leftrightarrow x^4+2x^3-3x^3-6x^2-x^2-2x+3x+6=0\)

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

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

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

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

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

b) \(2x^2+2xy+y^2+9=6x-\left|y+3\right|\)

\(\Leftrightarrow2x^2+2xy+y^2+9-6x+\left|y+3\right|=0\)

\(\Leftrightarrow\left(x^2+2xy+y^2\right)+x^2-6x+9+\left|y+3\right|=0\)

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

Do \(\left(x+y\right)^2\ge0;\left(x-3\right)^2\ge0;\left|y+3\right|\ge0\forall x;y\)

\(\Rightarrow\hept{\begin{cases}x+y=0\\x-3=0\\y+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=-3\end{cases}}\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}2x^2+x-3=0\\2x^2+x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2+2.x.\frac{1}{4}+\frac{1}{16}-\frac{1}{16}-\frac{3}{2}=0\\x^2+2.x.\frac{1}{4}+\frac{1}{16}-\frac{1}{16}-\frac{1}{2}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x+\frac{1}{4}\right)^2-\frac{25}{16}=0\\\left(x+\frac{1}{4}\right)^2-\frac{9}{16}=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}\left(x+\frac{1}{4}\right)^2=\frac{25}{16}\\\left(x+\frac{1}{4}\right)^2=\frac{9}{16}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{4}=\pm\frac{5}{4}\\x+\frac{1}{4}=\pm\frac{3}{4}\end{cases}}\)

Từ đó tính đc x

d) \(\left(x^2+3x+2\right)\left(x^2+7x+12\right)=24\)

\(\Leftrightarrow\left(x^2+x+2x+2\right)\left(x^2+3x+4x+12\right)=24\)

\(\Leftrightarrow\left[x\left(x+1\right)+2\left(x+1\right)\right]\left[x\left(x+3\right)+4\left(x+3\right)\right]=24\)

\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24=0\)

\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=0\)

Đặt \(x^2+5x+5=a\), khi đó pt có dạng:

\(\left(a-1\right)\left(a+1\right)-24=0\Leftrightarrow a^2-1-24=0\)

\(\Leftrightarrow a^2-25=0\Leftrightarrow\left(a-5\right)\left(a+5\right)=0\Leftrightarrow\orbr{\begin{cases}a=5\\a=-5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x^2+5x+5=5\\x^2+5x+5=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x\left(x+5\right)=0\\x^2+5x+10=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x\left(x+5\right)=0\\x^2+2.x.\frac{5}{2}+\frac{25}{4}+\frac{15}{4}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x\left(x+5\right)=0\\\left(x+\frac{5}{4}\right)^2=-\frac{15}{4}\left(vn\right)\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-5\end{cases}}\)

Đề mik ko hiểu lắm

\(2x\left(x-4\right)-6x^2\left(4-x\right)=0\)

\(\Leftrightarrow6x^2\left(x-4\right)+2x\left(x-4\right)=0\)

\(\Leftrightarrow2x\left(x-4\right)\left(3x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-\dfrac{1}{3}\end{matrix}\right.\)

\(1,\left(x-2\right)\left(x+2\right)\left(x^2+4\right)-\left(x^2-3\right)\left(x^2+3\right)\)

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

\(=x^2-16-x^2+9\)

\(=-7\)

\(2,\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(1+6x\right)\left(6x-1\right)\)

\(=\left(6x+1-6x+1\right)^2\)

\(=2^2=4\)

một đòn bẫy dài một mét .đặt ở đâu để có thể dùng 3600n có thể nâng tảng đá nặng 120kg?