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\(\dfrac{3^8\cdot20^5-3^9\cdot5^5\cdot2^9}{6^8\cdot10^4-3^8\cdot2^9\cdot5^4}=\dfrac{3^8\cdot2^{10}\cdot5^5-3^9\cdot5^5\cdot2^9}{2^8\cdot3^8\cdot2^4\cdot5^4-3^8\cdot2^9\cdot5^4}\\ =\dfrac{3^8\cdot2^9\cdot5^5\left(2-3\right)}{2^9\cdot3^8\cdot5^4\left(2^3-1\right)}=\dfrac{-5}{2^3-1}=\dfrac{-5}{7}\)
Áp dụng t/c dtsbn
\(\dfrac{x+1}{2}=\dfrac{y+2}{3}=\dfrac{z+2}{4}=\dfrac{3x+3-2y-4+z+2}{6-6+4}=\dfrac{-105+1}{4}=\dfrac{-104}{4}=-26\\ \Rightarrow\left\{{}\begin{matrix}x+1=-52\\y+2=-78\\z+2=-104\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-54\\y=-80\\z=-106\end{matrix}\right.\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x+1}{2}=\dfrac{y+2}{3}=\dfrac{z+2}{4}=\dfrac{3x-2y+z+3-4+2}{6-6+4}=\dfrac{-105+1}{4}=-26\)
Do đó: \(\left\{{}\begin{matrix}x+1=-52\\y+2=-78\\z+2=-104\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-53\\y=-80\\z=-106\end{matrix}\right.\)
a, 3 - 2 | 5x - 4 | = -11
2|5x - 4| = 14
|5x - 4| = 7
Th1: 5x -4 =7
5x = 11
x= 11/5
Th2:
5x -4 =-7
5x = -3
x= -3/5
a) => 2/5x-4/=14
=> /5x-4/=7
=> 5x-4=7 hoac 5x-4=-7
x=11/5 x=-3/5
Bài 9:
a: \(A=-0.5x^2yz\cdot\left(-3\right)xy^3z=1.5x^3y^4z^2\)
b: Hệ số là 1,5
Bậc là 9
a) \(\left(\frac{3}{5}x-\frac{2}{3}x-x\right).\frac{1}{7}=\frac{-5}{21}\)
\(\Rightarrow\left(\frac{3}{5}-\frac{2}{3}-1\right).x=\frac{-5}{21}:\frac{1}{7}=\frac{-5}{3}\)
\(\Rightarrow\frac{-16}{15}.x=\frac{-5}{3}\Rightarrow x=\frac{-5}{3}:\frac{-16}{15}=\frac{25}{16}\)
b) \(\left(x-\frac{1}{4}\right)^2=\frac{1}{36}\)
\(\Rightarrow\left(x-\frac{1}{4}\right)^2=\left(±\frac{1}{6}\right)^2\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{4}=\frac{1}{6}\\x-\frac{1}{4}=\frac{-1}{6}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{5}{12}\\x=\frac{1}{12}\end{cases}}\)
\(\left(3-\frac{2}{3}+\frac{4}{3}\right):\left(2\frac{1}{3}-2,5\right)^2\)
\(=\left(\frac{7}{3}+\frac{4}{3}\right):\left(2\frac{1}{3}-2\frac{1}{2}\right)^2\)
\(=\frac{11}{3}:\left(-\frac{1}{6}\right)^2\)
\(=\frac{11}{3}:\frac{1}{36}\)
\(=\frac{11}{3}x\frac{36}{1}\)
\(=\frac{396}{3}\)
\(=132\)
a) P(x) = -2x^2 + 4x^4 – 9x^3 + 3x^2 – 5x + 3
=4x^4-9x^3+x^2-5x+3
Q(x) = 5x^4 – x^3 + x^2 – 2x^3 + 3x^2 – 2 – 5x
=5x^4-3x^3+4x^2-5x-2
b)
P(x)
-bậc:4
-hệ số tự do:3
-hệ số cao nhất:4
Q(x)
-bậc :4
-hệ số tự do :-2
-hệ số cao nhất:5
bài 1)
a) \(\dfrac{\left(-3\right)^{10}.15^5}{25^3.\left(-9\right)^7}\)
\(=\dfrac{\left(-3\right)^{10}.\left(3.5\right)^5}{\left(5^2\right)^3.\left(-3.3\right)^7}\)
\(=\dfrac{\left(-3\right)^{10}.3^5.5^5}{5^6.\left(-3\right)^7.3^7}\)
\(=\dfrac{\left(-3\right)^3.1.1}{5.1.3^2}\)
\(=\dfrac{-27.1.1}{5.1.9}\)
\(=\dfrac{-27}{45}\)
\(=\dfrac{-9}{15}\)
b)\(2^3+3.\left(\dfrac{1}{9}\right)^0-2^{-2}.4\left[\left(-2\right)^2:\dfrac{1}{2}\right].8\)
\(=8+3.1-\dfrac{1}{2^2}.4+\left[\left(4:\dfrac{1}{2}\right)\right].8\)
\(=8+3.1-\dfrac{1}{4}.4+\left[4.\dfrac{2}{1}\right].8\)
\(=8+3.1-\dfrac{1}{4}.4+8.8\)
\(=8+3-1+64\)
\(=11-1+64\)
\(=10+64\)
\(=74\)
\(\dfrac{3}{4}:\left(2\dfrac{4}{9}\right)-\left|-3x+2\dfrac{2}{3}\right|=\dfrac{3}{4}\)
\(\Leftrightarrow\left|-3x+\dfrac{8}{3}\right|=\dfrac{3}{4}-\dfrac{3}{4}\cdot\dfrac{9}{22}\)
\(\Leftrightarrow\left|3x-\dfrac{8}{3}\right|=\dfrac{3}{4}-\dfrac{27}{88}=\dfrac{39}{88}\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-\dfrac{8}{3}=\dfrac{39}{88}\\3x-\dfrac{8}{3}=-\dfrac{39}{88}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{821}{792}\\x=\dfrac{587}{792}\end{matrix}\right.\)