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Bài 4:
a) \(\dfrac{4}{3}+\left(1,25-x\right)=2,25\)
\(1,25-x=2,25-\dfrac{4}{3}=\dfrac{9}{4}-\dfrac{4}{3}\)
\(1,25-x=\dfrac{11}{12}\)
\(x=1,25-\dfrac{11}{12}=\dfrac{5}{4}-\dfrac{11}{12}\)
\(x=\dfrac{1}{3}\)
b) \(\dfrac{17}{6}-\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(x-\dfrac{7}{6}=\dfrac{17}{6}-\dfrac{7}{4}=\dfrac{34}{12}-\dfrac{21}{12}\)
\(x-\dfrac{7}{6}=\dfrac{13}{12}\)
\(x=\dfrac{13}{12}+\dfrac{7}{6}=\dfrac{13}{12}+\dfrac{14}{12}\)
\(x=\dfrac{27}{12}=\dfrac{9}{4}\)
c) \(4-\left(2x+1\right)=3-\dfrac{1}{3}=\dfrac{9}{3}-\dfrac{1}{3}\)
\(4-\left(2x+1\right)=\dfrac{8}{3}\)
\(2x+1=\dfrac{8}{3}+4=\dfrac{8}{3}+\dfrac{12}{3}\)
\(2x+1=\dfrac{20}{3}\)
\(2x=\dfrac{20}{3}-1=\dfrac{20}{3}-\dfrac{3}{3}\)
\(2x=\dfrac{17}{3}\)
\(x=\dfrac{17}{3}.\dfrac{1}{2}=\dfrac{17}{6}\)
Bài 15:
a) \(\left(\dfrac{-2}{3}\right)^9:x=\dfrac{-2}{3}\)
\(x=\left(\dfrac{-2}{3}\right)^9:\dfrac{-2}{3}=\left(\dfrac{-2}{3}\right)^{9-1}\)
\(=>x=\left(\dfrac{-2}{3}\right)^8\)
b) \(x:\left(\dfrac{4}{9}\right)^5=\left(\dfrac{4}{9}\right)^4\)
\(x=\left(\dfrac{4}{9}\right)^4.\left(\dfrac{4}{9}\right)^5=\left(\dfrac{4}{9}\right)^{4+5}\)
\(=>x=\left(\dfrac{4}{9}\right)^9\)
c) \(\left(x+4\right)^3=-125\)
\(\left(x+4\right)^3=\left(-5\right)^3\)
\(=>x+4=-5\)
\(x=-5-4\)
\(=>x=-9\)
d) \(\left(10-5x\right)^3=64\)
\(\left(10-5x\right)^3=4^3\)
\(=>10-5x=4\)
\(5x=10-4\)
\(5x=6\)
\(=>x=\dfrac{6}{5}\)
e) \(\left(4x+5\right)^2=81\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(4x+5\right)^2=\left(-9\right)^2\\\left(4x+5\right)^2=9^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+5=-9\\4x+5=9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=-14\\4x=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-14}{4}\\x=1\end{matrix}\right.\)
Bài 16:
a) \(4-1\dfrac{2}{5}-\dfrac{8}{3}\)
\(=4-\dfrac{7}{5}-\dfrac{8}{3}\)
\(=\dfrac{60-21-40}{15}=\dfrac{-1}{15}\)
b) \(-0,6-\dfrac{-4}{9}-\dfrac{16}{15}\)
\(=\dfrac{-3}{5}+\dfrac{4}{9}-\dfrac{16}{15}\)
\(=\dfrac{\left(-27\right)+20-48}{45}=\dfrac{-55}{45}=\dfrac{-11}{9}\)
c) \(-\dfrac{15}{4}.\left(\dfrac{-7}{15}\right).\left(-2\dfrac{2}{5}\right)\)
\(=\dfrac{7}{4}.\dfrac{-12}{5}\)
\(=\dfrac{-21}{5}\)
\(#Wendy.Dang\)
a: \(=\dfrac{-3}{8}x^5\)
Hệ số là -3/8
Bậc là 5
b: \(=4x^2\cdot\dfrac{1}{8}=\dfrac{1}{2}\cdot x^2\)
Hệ sốlà 1/2
Bậc là 2
c: \(=\dfrac{16}{25}x^2\cdot\dfrac{15}{2}x^3=\dfrac{24}{5}x^5\)
Hệ số là 24/5
Bậc là 5
d: \(=\dfrac{2}{9}\cdot\dfrac{9}{4}\cdot x^2=\dfrac{1}{2}x^2\)
Hệ số là 1/2
Bậc là 2
e: \(=\dfrac{9}{25}x^2\cdot\dfrac{25}{4}x^6=\dfrac{9}{4}x^8\)
Hệ số là 9/4
Bậc là 8
Mấy bài dễ tự làm nhé:D
1)
Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{a}{a+b}=\dfrac{bk}{bk+b}=\dfrac{bk}{b\left(k+1\right)}=\dfrac{k}{k+1}\\\dfrac{c}{c+d}=\dfrac{dk}{dk+d}=\dfrac{dk}{d\left(k+1\right)}=\dfrac{k}{k+1}\end{matrix}\right.\)
Ta có điều phải chứng minh
\(\left\{{}\begin{matrix}\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{bk}{b\left(k-1\right)}=\dfrac{k}{k-1}\\\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{dk}{d\left(k-1\right)}=\dfrac{k}{k-1}\end{matrix}\right.\)
Ta có điều phải chứng minh
a.\(\dfrac{2}{3}\)x +1=\(\dfrac{7}{15}\)
\(\dfrac{2}{3}x\) =\(\dfrac{7}{15}-\dfrac{15}{15}\)
\(\dfrac{2}{3}x\) =\(\dfrac{-8}{15}\)
\(x\) =\(\dfrac{-8}{15}:\dfrac{2}{3}\)
\(x\) = \(\dfrac{-4}{5}\)
c.\(\dfrac{x}{12}=\dfrac{5}{9}\)
➝\(x.9=12.5\)
➝\(x.9=60\)
➝\(x=\dfrac{60}{9}=\dfrac{20}{3}\)
2.
a) \(\dfrac{x-1}{4}=\dfrac{9}{x-1}\)
\(\Leftrightarrow\left(x-1\right)^2=9.4\)
\(\Leftrightarrow\left(x-1\right)^2=36\)
\(\Leftrightarrow\left(x-1\right)^2=\left[{}\begin{matrix}6^2\\-6^2\end{matrix}\right.\)
\(\Leftrightarrow x-1=\left[{}\begin{matrix}6\\-6\end{matrix}\right.\)
\(\Leftrightarrow x=\left[{}\begin{matrix}7\\-5\end{matrix}\right.\)
1. a. \(\dfrac{3}{-1}=\dfrac{-15}{5}\); \(\dfrac{-1}{3}=\dfrac{5}{-15}\)
\(\dfrac{3}{-15}\)= \(\dfrac{5}{-1}\); \(\dfrac{-15}{3}=\dfrac{-1}{5}\)
b. \(\dfrac{4}{-3}=\dfrac{-12}{9};\dfrac{-3}{4}=\dfrac{9}{-12}\)
\(\dfrac{4}{-12}=\dfrac{9}{-3};\dfrac{-12}{4}=\dfrac{-3}{9}\)
c. \(\dfrac{3}{7}=\dfrac{c}{b};\dfrac{7}{3}=\dfrac{b}{c}\)
\(\dfrac{3}{c}=\dfrac{b}{7};\dfrac{c}{3}=\dfrac{7}{b}\)
d. \(\dfrac{a}{b}=\dfrac{y}{x};\dfrac{b}{a}=\dfrac{x}{y}\)
\(\dfrac{a}{y}=\dfrac{x}{b};\dfrac{y}{a}=\dfrac{b}{x}\)
chúc bạn học tốt
C
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