d) \(2x^2+5x-7=0\)

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

bạn xem lại đề bài 1 là GTNN hay GTLN nha

\(a)x+30\%x=-1,31\)

\(\Leftrightarrow x+\frac{3x}{10}=-1,31\)

\(\Leftrightarrow10x+3x=-13,1\)

\(\Leftrightarrow13x=-13,1\Leftrightarrow x=-\frac{131}{130}\)

\(b)\left(x-\frac{1}{2}\right):\frac{1}{3}+\frac{5}{7}=9\frac{5}{7}\)

\(\Leftrightarrow\frac{2x-1}{2}.3+\frac{5}{7}=\frac{68}{7}\)

\(\Leftrightarrow\frac{6x-3}{2}=\frac{63}{7}\)

\(\Leftrightarrow\frac{6x-3}{2}=9\)

\(\Leftrightarrow6x-3=18\)

\(\Leftrightarrow x=\frac{7}{2}\)

\(c)\frac{1}{2}x-\frac{3}{4}=\frac{14}{9}.\frac{3}{7}\)

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

\(\Leftrightarrow\frac{x}{2}=\frac{17}{12}\)

\(\Leftrightarrow12x=34\Leftrightarrow x=\frac{17}{6}\)

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

a, 2.(4x-3)-3(x+5)+4(x-10)=5(x+2)

    2.4x-2.3-3.x+3.5+4x-4.10=5x+5.2

    8x-6-3x+15+4x-40=5x-10

    8x-3x+4x-5x-6-15-40-10=0

    4x-71=0

    4x=71

     x=71:4

    x=71/4

a: \(x-\dfrac{10}{3}=\dfrac{7}{15}\cdot\dfrac{3}{5}\)

=>\(x-\dfrac{10}{3}=\dfrac{21}{75}=\dfrac{7}{25}\)

=>\(x=\dfrac{7}{25}+\dfrac{10}{3}=\dfrac{21+250}{75}=\dfrac{271}{75}\)

b: \(x+\dfrac{3}{22}=\dfrac{27}{121}\cdot\dfrac{9}{11}\)

=>\(x+\dfrac{3}{22}=\dfrac{243}{1331}\)

=>\(x=\dfrac{243}{1331}-\dfrac{3}{22}=\dfrac{123}{2662}\)

c: \(\dfrac{8}{23}\cdot\dfrac{46}{24}-x=\dfrac{1}{3}\)

=>\(\dfrac{8}{24}\cdot\dfrac{46}{23}-x=\dfrac{1}{3}\)

=>\(\dfrac{2}{3}-x=\dfrac{1}{3}\)

=>\(x=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\)

d: \(1-x=\dfrac{49}{65}\cdot\dfrac{5}{7}\)

=>\(1-x=\dfrac{49}{7}\cdot\dfrac{5}{65}=\dfrac{7}{13}\)

=>\(x=1-\dfrac{7}{13}=\dfrac{6}{13}\)

a: \(x+7⋮x+2\)

=>\(x+2+5⋮x+2\)

=>\(5⋮x+2\)

=>\(x+2\in\left\{1;-1;5;-5\right\}\)

=>\(x\in\left\{-1;-3;3;-7\right\}\)

b: \(2x+5⋮x+1\)

=>\(2x+2+3⋮x+1\)

=>\(3⋮x+1\)

=>\(x+1\in\left\{1;-1;3;-3\right\}\)

=>\(x\in\left\{0;-2;2;-4\right\}\)

c: \(3x-2⋮x+3\)

=>\(3x+9-11⋮x+3\)

=>\(-11⋮x+3\)

=>\(x+3\in\left\{1;-1;11;-11\right\}\)

=>\(x\in\left\{-2;-4;8;-14\right\}\)

d: \(12x+1⋮3x+2\)

=>\(12x+8-7⋮3x+2\)

=>\(-7⋮3x+2\)

=>\(3x+2\in\left\{1;-1;7;-7\right\}\)

=>\(3x\in\left\{-1;-3;5;-9\right\}\)

=>\(x\in\left\{-\dfrac{1}{3};-1;\dfrac{5}{3};-3\right\}\)

e: \(x^2+3x+5⋮x+3\)

=>\(x\left(x+3\right)+5⋮x+3\)

=>\(5⋮x+3\)

=>\(x+3\in\left\{1;-1;5;-5\right\}\)

=>\(x\in\left\{-2;-4;2;-8\right\}\)

f: \(x^2-2x+3⋮x+2\)

=>\(x^2+2x-4x-8+11⋮x+2\)

=>\(11⋮x+2\)

=>\(x+2\in\left\{1;-1;11;-11\right\}\)

=>\(x\in\left\{-1;-3;9;-13\right\}\)

a) Ta có: \(\left(x-3\right)=\left(3-x\right)^2\)

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

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

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

b) Ta có: \(x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}=\dfrac{1}{64}\)

\(\Leftrightarrow x^3+3\cdot x^2\cdot\dfrac{1}{2}+3\cdot x\cdot\dfrac{1}{4}+\left(\dfrac{1}{2}\right)^3=\dfrac{1}{64}\)

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

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

hay \(x=-\dfrac{1}{4}\)

c) Ta có: \(8x^3-50x=0\)

\(\Leftrightarrow2x\left(4x^2-25\right)=0\)

\(\Leftrightarrow x\left(2x-5\right)\left(2x+5\right)=0\)

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

e) Ta có: \(x\left(x+3\right)-x^2-3x=0\)

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

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

f) Ta có: \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)

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

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

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