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b, Ta có : \(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{5}=\dfrac{z}{6}\Rightarrow\dfrac{x}{15}=\dfrac{y}{20}=\dfrac{z}{24}\)
Đặt \(x=15k;y=20k;z=24k\)
Thay vào A ta được : \(A=\dfrac{30k+60k+96k}{45k+80k+120k}=\dfrac{186k}{245k}=\dfrac{186}{245}\)
Bài 1:
b) ĐKXĐ: \(x\ne3\)
Ta có: \(\dfrac{3-x}{20}=\dfrac{-5}{x-3}\)
\(\Leftrightarrow\dfrac{x-3}{-20}=\dfrac{-5}{x-3}\)
\(\Leftrightarrow\left(x-3\right)^2=100\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=10\\x-3=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=13\left(nhận\right)\\x=-7\left(nhận\right)\end{matrix}\right.\)
Vậy: \(x\in\left\{13;-7\right\}\)
Bài 1:
|\(x\)| = 1 ⇒ \(x\) \(\in\) {-\(\dfrac{1}{3}\); \(\dfrac{1}{3}\)}
A(-1) = 2(-\(\dfrac{1}{3}\))2 - 3.(-\(\dfrac{1}{3}\)) + 5
A(-1) = \(\dfrac{2}{9}\) + 1 + 5
A (-1) = \(\dfrac{56}{9}\)
A(1) = 2.(\(\dfrac{1}{3}\) )2- \(\dfrac{1}{3}\).3 + 5
A(1) = \(\dfrac{2}{9}\) - 1 + 5
A(1) = \(\dfrac{38}{9}\)
|y| = 1 ⇒ y \(\in\) {-1; 1}
⇒ (\(x;y\)) = (-\(\dfrac{1}{3}\); -1); (-\(\dfrac{1}{3}\); 1); (\(\dfrac{1}{3};-1\)); (\(\dfrac{1}{3};1\))
B(-\(\dfrac{1}{3}\);-1) = 2.(-\(\dfrac{1}{3}\))2 - 3.(-\(\dfrac{1}{3}\)).(-1) + (-1)2
B(-\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\) - 1 + 1
B(-\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\)
B(-\(\dfrac{1}{3}\); 1) = 2.(-\(\dfrac{1}{3}\))2 - 3.(-\(\dfrac{1}{3}\)).1 + 12
B(-\(\dfrac{1}{3};1\)) = \(\dfrac{2}{9}\) + 1 + 1
B(-\(\dfrac{1}{3}\); 1) = \(\dfrac{20}{9}\)
B(\(\dfrac{1}{3};-1\)) = 2.(\(\dfrac{1}{3}\))2 - 3.(\(\dfrac{1}{3}\)).(-1) + (-1)2
B(\(\dfrac{1}{3}\); -1) = \(\dfrac{2}{9}\) + 1 + 1
B(\(\dfrac{1}{3}\); -1) = \(\dfrac{20}{9}\)
B(\(\dfrac{1}{3}\); 1) = 2.(\(\dfrac{1}{3}\))2 - 3.(\(\dfrac{1}{3}\)).1 + (1)2
B(\(\dfrac{1}{3}\); 1) = \(\dfrac{2}{9}\) - 1 + 1
B(\(\dfrac{1}{3}\);1) = \(\dfrac{2}{9}\)
Bài 2:
a: \(=2x^4-x^3-10x^2-2x^3+x^2+10x=2x^3-3x^3-9x^2+10x\)
b: \(=\left(x^2-15x\right)\left(x^2-7x+3\right)\)
\(=x^4-7x^3+3x^2-15x^3+105x^2-45x\)
\(=x^4-22x^3+108x^2-45x\)
c: \(=12x^5-18x^4+30x^3-24x^2\)
d: \(=-3x^6+2.4x^5-1.2x^4+1.8x^2\)
a, \(\left|2x-3\right|-\dfrac{1}{3}=0\Leftrightarrow\left|2x-3\right|=\dfrac{1}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=\dfrac{1}{3}\\2x-3=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\)
b, tương tự
c, \(\left|2x-1\right|-\left|x+\dfrac{1}{3}\right|=0\Leftrightarrow\left|2x-1\right|=\left|x+\dfrac{1}{3}\right|\)
TH1 : \(2x-1=x+\dfrac{1}{3}\Leftrightarrow x=\dfrac{4}{3}\)
TH2 : \(2x-1=-x-\dfrac{1}{3}\Leftrightarrow3x=\dfrac{2}{3}\Leftrightarrow x=\dfrac{2}{9}\)
d, \(3x-\left|x+15\right|=\dfrac{5}{4}\Leftrightarrow\left|x+15\right|=3x-\dfrac{5}{4}\)ĐK : x >= 5/12
TH1 : \(x+15=3x-\dfrac{5}{4}\Leftrightarrow-2x=-\dfrac{65}{4}\Leftrightarrow x=\dfrac{65}{8}\)( tm )
TH2 : \(x+15=\dfrac{5}{3}-3x\Leftrightarrow4x=-\dfrac{40}{3}\Leftrightarrow x=-\dfrac{10}{3}\)
a: \(A\left(x\right)+B\left(x\right)\)
\(=-2x^3+11x^2-5x-\dfrac{1}{5}+2x^3-3x^2-7x+\dfrac{1}{5}\)
\(=8x^2-12x\)
b: C(x)=A(x)-B(x)
\(=-2x^3+11x^2-5x-\dfrac{1}{5}-2x^3+3x^2+7x-\dfrac{1}{5}\)
\(=-4x^3+14x^2+2x-\dfrac{2}{5}\)
câu 1.
đặt A=\(\dfrac{15}{11.14}+\dfrac{15}{14.17}+...+\dfrac{15}{65.68}+\dfrac{15}{68.71}\)
xét \(\dfrac{A}{3}\)=\(\dfrac{15}{3.11.14}+\dfrac{15}{3.14.17}+...+\dfrac{15}{3.65.68}+\dfrac{15}{3.68.71}\)
ta có:+ \(\dfrac{15}{3.11.14}=\dfrac{15}{3}\left(\dfrac{1}{11}-\dfrac{1}{14}\right)=\dfrac{15}{3.11}-\dfrac{15}{3.14}\)
tương tự ta có:
+\(\dfrac{15}{3.11.14}=\dfrac{15}{3.11}-\dfrac{15}{3.14}\)
+\(\dfrac{15}{3.14.17}=\dfrac{15}{3.14}-\dfrac{15}{3.17}\)
....
+\(\dfrac{15}{3.65.68}=\dfrac{15}{3.65}-\dfrac{15}{3.68}\)
+\(\dfrac{15}{3.68.71}=\dfrac{15}{3.68}-\dfrac{15}{3.71}\)
cộng vế theo vế ta đc:
\(\dfrac{15}{3.11.14}+\dfrac{15}{3.14.17}+...+\dfrac{15}{3.65.68}+\dfrac{15}{3.68.71}\)
=\(\dfrac{15}{3.11}-\dfrac{15}{3.14}+\dfrac{15}{3.14}-\dfrac{15}{3.17}+...+\dfrac{15}{3.65}-\dfrac{15}{3.68}+\dfrac{15}{3.68}-\dfrac{15}{3.71}=\dfrac{15}{3.11}-\dfrac{15}{3.71}\)
=> \(\dfrac{A}{3}\)=\(\dfrac{15}{3.11}-\dfrac{15}{3.71}\)
=> A= \(\dfrac{15}{11}-\dfrac{15}{17}=\dfrac{90}{187}\)
câu 1b.
trước khi làm bài này có chú ý này:\(0^n=0\)với n\(\ne0\) và \(a^0=1\)với a\(\ne0\)
đặt: \(t=\left(x-5\right)\Rightarrow\left\{{}\begin{matrix}\left(x-5\right)^{x+1}=\left(x-5\right)^{x-5+6}=t^{t+6}\\\left(x-5\right)^{x+2015}=\left(x-5\right)^{x-5+2020}=t^{t+2020}\end{matrix}\right.\)
=> \(\left(x-5\right)^{x+1}-\left(x-5\right)^{x+2015}=0\)
\(\Leftrightarrow\)\(t^{t+6}-t^{t+2020}=0\Leftrightarrow t^{t+6}\left(1-t^{2014}\right)=0\Leftrightarrow\left[{}\begin{matrix}t^{t+6}=0^{t+6}\\1-t^{2014}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=0\\t^{2014}=1=1^{2014}\Rightarrow t=1\end{matrix}\right.\)với t=0 => x-5=0=> x=5
với t=1=> x-5=1=>x=6