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

\(\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\5-x^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\-x^2=-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=\pm\sqrt{5}\end{matrix}\right.\)

Vậy \(S=\left\{-\dfrac{2}{3};-\sqrt{5};\sqrt{5}\right\}\)

\(2,-2x-\dfrac{2}{3}\left(\dfrac{3}{4}-\dfrac{1}{8}x\right)=\left(-\dfrac{1}{2}\right)^3\)

\(\Leftrightarrow-2x-\dfrac{1}{2}+\dfrac{1}{12}x=-\dfrac{1}{8}\)

\(\Leftrightarrow-2x+\dfrac{1}{12}x=-\dfrac{1}{8}+\dfrac{1}{2}\)

\(\Leftrightarrow-\dfrac{23}{12}=\dfrac{3}{8}\)

\(\Leftrightarrow x=-\dfrac{9}{46}\)

Vậy \(S=\left\{-\dfrac{9}{46}\right\}\)

\(3,\dfrac{1}{12}:\dfrac{4}{21}=3\dfrac{1}{2}:\left(3x-2\right)\)

\(\Leftrightarrow\dfrac{1}{12}.\dfrac{21}{4}=\dfrac{7}{2}.\dfrac{1}{3x-2}\)

\(\Leftrightarrow\dfrac{7}{16}=\dfrac{7}{6x-4}\)

\(\Leftrightarrow6x-4=7:\dfrac{7}{16}\)

\(\Leftrightarrow6x-4=16\)

\(\Leftrightarrow x=\dfrac{10}{3}\)

Vậy \(S=\left\{\dfrac{10}{3}\right\}\)

\(4,\dfrac{x-1}{x+2}=\dfrac{4}{5}\left(dk:x\ne-2\right)\)

\(\Rightarrow5\left(x-1\right)=4\left(x+2\right)\)

\(\Rightarrow5x-5=4x+8\)

\(\Rightarrow x=13\left(tmdk\right)\)

Vậy \(S=\left\{13\right\}\)

\(a,\Leftrightarrow\left[{}\begin{matrix}-\dfrac{4}{3}x+\dfrac{1}{2}=\dfrac{1}{2}\\-\dfrac{4}{3}x+\dfrac{1}{2}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{4}\end{matrix}\right.\\ c,\Leftrightarrow\left(\dfrac{1}{2}\right)^x\left(1+\dfrac{1}{4}\right)=\dfrac{5}{4}\\ \Leftrightarrow\left(\dfrac{1}{2}\right)^x=1\Leftrightarrow x=0\)

b: Ta có: \(3^x+3^{x+2}=20\)

\(\Leftrightarrow3^x\cdot10=20\)

\(\Leftrightarrow3^x=2\left(loại\right)\)

1/1 x 2 + 1/2 x 3 + 1/3 x 4 + ... + 1/999 x 1000 + 1

= 1/1 - 1/1000 + 1 

= 999/1000 + 1

= 1999/1000

Chuc ban may man

1,999 nha

a) \(\sqrt{4x^2+4x+1}=6\)

\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

b) \(\sqrt{4x^2-4\sqrt{7}x+7}=\sqrt{7}\)

\(\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)

\(\Leftrightarrow\left(2x-\sqrt{7}\right)^2=\left(\sqrt{7}\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt[]{7}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)

a) \(\sqrt{4x^2+4x+1}=6\)

\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left|2x+1\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

b) \(pt\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)

\(\Leftrightarrow\left|2x-\sqrt{7}\right|=\sqrt{7}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)

c) \(PT\Leftrightarrow\sqrt{\left(x+\sqrt{3}\right)^2}=2\sqrt{3}\)

\(\Leftrightarrow\left|x+\sqrt{3}\right|=2\sqrt{3}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\sqrt{3}=2\sqrt{3}\\x+\sqrt{3}=-2\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}\\x=-3\sqrt{3}\end{matrix}\right.\)

d) \(pt\Leftrightarrow\left|x-3\right|=9\Leftrightarrow\left[{}\begin{matrix}x-3=-9\\x-3=9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=12\end{matrix}\right.\)

 1 + ( 1 + 2 ) + ( 1 + 2 + 3 ) + ( 1 + 2 + 3 + 4 ) + ……+ ( 1 + 2 + 3 +…+ 99 ) = x

Ta thấy :  số 1 xuất hiện trong 99 tổng ,  số 2 xuất hiện trong 98 lần , số 3 xuất hiện trong 97 tổng , ... , 99 xuất hiện trong 1 tổng 

Nên tổng trên bằng ; 1 x 99 + 2 x 98 + 3 x 97 + ... + 97 x 3 + 98 x 2 + 99 x 1 = x 

                               [( 1 x99 ) +  ( 99 x1 )] + [( 2 x 98 ) + ( 98 x 2 ) ] + ... + [( 49 x 51 ) + ( 51 x 49 )] = x 

 ( Tự làm tiếp )

làm đi hộ mk mk k 5 k cho

\(5\frac{9}{10}:\frac{3}{2}-\left(2\frac{1}{3}x4\frac{1}{2}-2\frac{1}{2}\right):\frac{7}{4}\)

=\(\frac{59}{10}:\frac{3}{2}-\left(\frac{7}{3}x\frac{9}{2}-\frac{5}{2}\right):\frac{7}{4}\)

=\(\frac{118}{30}-\left(\frac{63}{6}-\frac{5}{2}\right)x\frac{4}{7}\)

=\(\frac{118}{30}-\left(\frac{63}{6}-\frac{15}{6}\right)x\frac{4}{7}\)

=\(\frac{118}{30}-8x\frac{4}{7}\)

=\(\frac{118}{30}-\frac{32}{7}\)

sau bn tự tính nha