Giải:

\(\dfrac{1}{2}.\left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2-\dfrac{1}{5}=-\dfrac{3}{40}\)

\(\Leftrightarrow\dfrac{1}{2}.\left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2=-\dfrac{3}{40}+\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{1}{2}.\left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2=\dfrac{1}{8}\)

\(\Leftrightarrow\left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2=\dfrac{1}{8}:\dfrac{1}{2}\)

\(\Leftrightarrow\left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2=\dfrac{1}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{3}x-\dfrac{1}{5}=\dfrac{1}{2}\\\dfrac{1}{3}x-\dfrac{1}{5}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{3}x=\dfrac{7}{10}\\\dfrac{1}{3}x=-\dfrac{3}{10}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{21}{10}\\x=-\dfrac{9}{10}\end{matrix}\right.\)

Vậy ...

Chúc bạn học tốt!

\(\dfrac{1}{2}.\left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2-\dfrac{1}{5}=-\dfrac{3}{40}\\ \dfrac{1}{2}\left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2=-\dfrac{3}{40}+\dfrac{1}{5}\\ \dfrac{1}{2}\left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2=\dfrac{1}{8}\\ \left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2=\dfrac{1}{8}:\dfrac{1}{2}\\\left(\dfrac{1}{3}x-\dfrac{1}{5}\right)^2=\dfrac{1}{4}\\ \left(\dfrac{1}{3}x-\dfrac{1}{5}\right)=\left(\pm\dfrac{1}{2}\right)^2\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{1}{3}x-\dfrac{1}{5}=\dfrac{1}{2}\\\dfrac{1}{3}x-\dfrac{1}{5}=-\dfrac{1}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{3}x=\dfrac{7}{10}\\\dfrac{1}{3}x=\dfrac{3}{10}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{21}{10}\\x=\dfrac{9}{10}\end{matrix}\right. \)

Vậy \(x=\dfrac{21}{10}\) hoặc \(x=\dfrac{9}{10}\)

a)

Đặt

\(\sqrt{1+x}=a; \sqrt{1-x}=b\Rightarrow \left\{\begin{matrix} ab=\sqrt{(1+x)(1-x)}=\sqrt{1-x^2}\\ a\geq b\\ a^2+b^2=2\end{matrix}\right.\)

Khi đó:

\(A=\frac{\sqrt{1-\sqrt{1-x^2}}(\sqrt{(1+x)^3}+\sqrt{(1-x)^3})}{2-\sqrt{1-x^2}}\)

\(=\frac{\sqrt{\frac{a^2+b^2}{2}-ab}(a^3+b^3)}{a^2+b^2-ab}=\frac{\sqrt{\frac{a^2+b^2-2ab}{2}}(a+b)(a^2-ab+b^2)}{a^2+b^2-ab}\)

\(=\sqrt{\frac{a^2-2ab+b^2}{2}}(a+b)=\sqrt{\frac{(a-b)^2}{2}}(a+b)=\frac{1}{\sqrt{2}}|a-b|(a+b)\)

\(=\frac{1}{\sqrt{2}}(a-b)(a+b)=\frac{1}{\sqrt{2}}(a^2-b^2)=\frac{1}{\sqrt{2}}[(1+x)-(1-x)]=\sqrt{2}x\)

Sửa đề: \(\frac{25}{(x+z)^2}=\frac{16}{(z-y)(2x+y+z)}\)

Ta có:

Áp dụng tính chất dãy tỉ số bằng nhau thì:

\(k=\frac{a}{x+y}=\frac{5}{x+z}=\frac{a+5}{2x+y+z}=\frac{5-a}{z-y}\) ($k$ là một số biểu thị giá trị chung)

Khi đó:

\(\frac{16}{(z-y)(2x+y+z)}=\frac{25}{(x+z)^2}=(\frac{5}{x+z})^2=k^2\)

Mà: \(k^2=\frac{a+5}{2x+y+z}.\frac{5-a}{z-y}=\frac{25-a^2}{(2x+y+z)(z-y)}\)

Do đó: \(\frac{16}{(z-y)(2x+y+z)}=\frac{25-a^2}{(2x+y+z)(z-y)}\Rightarrow 16=25-a^2\)

\(\Rightarrow a^2=9\Rightarrow a=\pm 3\)

Suy ra:
\(Q=\frac{a^6-2a^5+a-2}{a^5+1}=\frac{a^5(a-2)+(a-2)}{a^5+1}=\frac{(a-2)(a^5+1)}{a^5+1}=a-2=\left[\begin{matrix} 1\\ -5\end{matrix}\right.\)

d) \(\left(x+2\right)\left(x^2-2x+4\right)\)

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

\(=x^3+2^3\)

\(=x^3+8\)

e) \(\left(\dfrac{1}{4}-\dfrac{x}{5}\right)\left(\dfrac{x^2}{25}+\dfrac{x}{20}+\dfrac{1}{16}\right)\)

\(=\left(\dfrac{1}{4}-\dfrac{1}{5}x\right)\left(\dfrac{1}{25}x^2+\dfrac{1}{5}x\cdot\dfrac{1}{4}+\dfrac{1}{16}\right)\)

\(=\left(\dfrac{1}{4}-\dfrac{1}{5}x\right)\left[\left(\dfrac{1}{5}x\right)^2+\dfrac{1}{5}x\cdot\dfrac{1}{4}+\left(\dfrac{1}{4}\right)^2\right]\)

\(=\left(\dfrac{1}{4}\right)^3-\left(\dfrac{1}{5}x\right)^3\)

\(=\dfrac{1}{64}-\dfrac{1}{125}x^3\)

\(=\dfrac{1}{64}-\dfrac{x^3}{125}\)

d: (x+2)(x^2-2x+4)

=(x+2)(x^2-x*2+2^2)

=x^3+8

e: (1/4-x/5)(1/16+x/20+x^2/25)

=(1/4-x/5)[(1/4)^2+1/4*x/5+(x/5)^2]

=1/64-x^3/125

a) \(\left(\dfrac{x^2}{2}+y^2\right)^2\)

\(=\left(\dfrac{1}{2}x^2+y^2\right)^2\)

\(=\left(\dfrac{1}{2}x^2\right)^2+2\cdot\dfrac{1}{2}x^2\cdot y^2+\left(y^2\right)^2\)

\(=\dfrac{1}{4}x^4+x^2y^2+y^4\)

b) \(\left(\dfrac{4}{5}x^2-\dfrac{2}{3}y\right)^2\)

\(=\left(\dfrac{4}{5}x^2\right)^2-2\cdot\dfrac{4}{5}x^2\cdot\dfrac{2}{3}y+\left(\dfrac{2}{3}y\right)^2\)

\(=\dfrac{16}{25}x^4-\dfrac{16}{15}x^2y+\dfrac{4}{9}y^2\)

c) \(\left(2x+\dfrac{1}{2}\right)\left(2x-\dfrac{1}{2}\right)\)

\(=\left(2x\right)^2-\left(\dfrac{1}{2}\right)^2\)

\(=4x^2-\dfrac{1}{4}\)

a: (1/2x^2+y^2)^2

=(1/2x^2)^2+2*1/2x^2*y^2+y^4

=1/4x^4+x^2y^2+y^4

b: (4/5x^2-2/3y)^2

=(4/5x^2)^2-2*4/5x^2*2/3y+4/9y^2

=16/25x^4-16/15x^2y+4/9y^2

c: =(2x)^2-(1/2)^2

=4x^2-1/4

\(-\dfrac{2}{5}+\dfrac{5}{3}\left(\dfrac{3}{2}-\dfrac{4}{15}x\right)=-\dfrac{7}{6}\)

\(\Rightarrow\dfrac{5}{3}\left(\dfrac{3}{2}-\dfrac{4}{15}x\right)=-\dfrac{23}{30}\)

\(\Rightarrow\dfrac{3}{2}-\dfrac{4}{15}x=-\dfrac{23}{50}\)

\(\Rightarrow\dfrac{4}{15}x=\dfrac{49}{25}\Rightarrow x=\dfrac{147}{20}\)

Chúc bạn học tốt!!!

a) \(\left(x-\dfrac{1}{2}\right)\left(-3-\dfrac{x}{2}\right)=0\)

Th1 : \(x-\dfrac{1}{2}=0\)

         \(x=0+\dfrac{1}{2}\)

         \(x=\dfrac{1}{2}\)

Th2 : \(-3-\dfrac{x}{2}=0\)

         \(\dfrac{x}{2}=-3\)

         \(x=\left(-3\right)\cdot2\)

         \(x=-6\)

Vậy \(x\) = \(\left(\dfrac{1}{2};-6\right)\)

b) \(x-\dfrac{1}{8}=\dfrac{5}{8}\)

    \(x=\dfrac{5}{8}+\dfrac{1}{8}\)

   \(x=\dfrac{3}{4}\)

c) \(-\dfrac{1}{2}-\left(\dfrac{3}{2}+x\right)=-2\)

                \(\dfrac{3}{2}+x=-\dfrac{1}{2}-\left(-2\right)\)

                \(\dfrac{3}{2}+x=\dfrac{3}{2}\)

                       \(x=\dfrac{3}{2}-\dfrac{3}{2}\)

                      \(x=0\)

d) \(x+\dfrac{1}{3}=\dfrac{-12}{5}\cdot\dfrac{10}{6}\)

    \(x+\dfrac{1}{3}=-4\)

    \(x=-4-\dfrac{1}{3}\)

    \(x=-\dfrac{13}{3}\)

a) \(\left|3x-\dfrac{1}{2}\right|+\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|=0\)

Do \(\left|3x-\dfrac{1}{2}\right|,\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|\ge0\forall x,y\)

\(\Rightarrow\left\{{}\begin{matrix}3x-\dfrac{1}{2}=0\\\dfrac{1}{4}y+\dfrac{3}{5}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=-\dfrac{12}{5}\end{matrix}\right.\)

b) \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|+\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\le0\)

Do \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|,\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\ge0\forall x,y\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x+\dfrac{1}{9}=0\\\dfrac{5}{7}y-\dfrac{1}{2}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{27}\\y=\dfrac{7}{10}\end{matrix}\right.\)

Bài 2:

\(\left\{{}\begin{matrix}\left(2x-\dfrac{1}{2}\right)^2\ge0\\\left(y+\dfrac{1}{2}\right)^2\ge0\\\left(z-\dfrac{1}{3}\right)^2\ge0\end{matrix}\right.\Rightarrow\left(2x-\dfrac{1}{2}\right)^2+\left(y+\dfrac{1}{2}\right)^2+\left(z-\dfrac{1}{3}\right)^2\ge0\)Mà \(\left(2x-\dfrac{1}{2}\right)^2+\left(y+\dfrac{1}{2}\right)^2+\left(z-\dfrac{1}{3}\right)^2=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(2x-\dfrac{1}{2}\right)^2=0\\\left(y+\dfrac{1}{2}\right)^2=0\\\left(z-\dfrac{1}{3}\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{4}\\y=\dfrac{-1}{2}\\z=\dfrac{1}{3}\end{matrix}\right.\)

Vậy \(x=\dfrac{1}{4},y=\dfrac{-1}{2},z=\dfrac{1}{3}\)

1)

a) \(2x+\dfrac{5}{2}=\dfrac{7}{2}\)

\(\Leftrightarrow2x=\dfrac{7}{2}-\dfrac{5}{2}\)

\(\Leftrightarrow2x=1\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

Vậy \(x=\dfrac{1}{2}\)

b) \(\left|5-\dfrac{1}{2}x\right|=\left|-\dfrac{1}{5}\right|\)

\(\Leftrightarrow\left|5-\dfrac{1}{2}x\right|=\dfrac{1}{5}\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{48}{5}\\x=\dfrac{52}{5}\end{matrix}\right.\)

Vậy \(x_1=\dfrac{48}{5};x_2=\dfrac{52}{5}\)

a/ \(2x+\dfrac{5}{2}=\dfrac{7}{2}\)

\(\Rightarrow2x=1\Rightarrow x=\dfrac{1}{2}\)

b/ \(\left|5-\dfrac{1}{2}x\right|=\left|-\dfrac{1}{5}\right|\)

\(\Rightarrow\left|5-\dfrac{1}{2}x\right|=\dfrac{1}{5}\)

\(\Rightarrow\left[{}\begin{matrix}5-\dfrac{1}{2}x=\dfrac{1}{5}\\5-\dfrac{1}{2}x=\dfrac{1}{5}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{12}{5}\\x=\dfrac{13}{5}\end{matrix}\right.\)

Bài 2:

Vì \(\left\{{}\begin{matrix}\left(2x-\dfrac{1}{2}\right)^2\ge0\forall x\\\left(y+\dfrac{1}{2}\right)^2\ge0\forall y\\\left(x-\dfrac{1}{3}\right)^2\ge0\forall z\end{matrix}\right.\)

=> Để bt = 0 thì\(\left\{{}\begin{matrix}\left(2x-\dfrac{1}{2}\right)^2=0\\\left(y+\dfrac{1}{2}\right)^2=0\\\left(z-\dfrac{1}{3}\right)^2=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2x-\dfrac{1}{2}=0\\y+\dfrac{1}{2}=0\\z-\dfrac{1}{3}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{4}\\y=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy.....................