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giải chi tiết giúp mk vs nha 
Bài 2:
a: \(\left|x\right|=-x\)
nên x<=0
b: \(\left|x\right|>x\)
=>x<0
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
Bài 1:
1: \(M=\left|x-1\right|+x+2\)
Trường hợp 1: x>=1
M=x-1+x+2=2x+1
Trường hợp 2: x<1
M=1-x+x+2=3
2: \(N=x-3+\left|x-3\right|\)
Trường hợp 1: x>=3
\(N=x-3+x-3=2x-6\)
Trường hợp 2: x<3
\(N=x-3+3-x=0\)
3: \(P=2x-1-\left|x-2\right|\)
Trường hợp 1: x<2
\(P=2x-1-\left(2-x\right)=2x-1-2+x=3x-3\)
TRường hợp 2: x>=2
\(P=2x-1-x+2=x+1\)
Bài 1:
a: \(=\dfrac{15-32}{40}\cdot10+\dfrac{1}{4}\)
\(=\dfrac{-17}{4}+\dfrac{1}{4}=-\dfrac{16}{4}=-4\)
b: \(=\left(\dfrac{9}{6}-\dfrac{5}{6}\right)^2+\dfrac{5}{2}+\dfrac{2}{3}\)
\(=\dfrac{4}{9}+\dfrac{5}{2}+\dfrac{2}{3}\)
\(=\dfrac{8}{18}+\dfrac{45}{18}+\dfrac{12}{18}=\dfrac{65}{18}\)
1.
a.
\(\left(\dfrac{-4}{5}+\dfrac{2}{3}\right)\cdot\dfrac{7}{11}+\left(\dfrac{-1}{5}+\dfrac{1}{3}\right)\cdot\dfrac{7}{11}\\ =\dfrac{7}{11}\cdot\left(\dfrac{-4}{5}+\dfrac{2}{3}+\dfrac{-1}{5}+\dfrac{1}{3}\right) \\ =\dfrac{7}{11}\cdot\left[\left(\dfrac{-4}{5}+\dfrac{-1}{5}\right)+\left(\dfrac{1}{3}+\dfrac{2}{3}\right)\right]\\ =\dfrac{7}{11}\cdot\left[\left(-1\right)+1\right]\\ =\dfrac{7}{11}\cdot0\\ =0\)
b.
\(\left(-3^2\right)\cdot\left(\dfrac{3}{4}-0,25\right)-\left|-2\right|\\ =\left(-9\right)\cdot0,5-2\\ =-4,5-2\\ =-6,5\)
2.
\(y=f\left(x\right)=\left(m+1\right)x\\ \Rightarrow4=f\left(2\right)=\left(m+1\right)\cdot2\\ \Rightarrow m+1=2\\ \Leftrightarrow m=1\)
Tự
3.
a.
\(\left|x-\dfrac{2}{5}\right|=\dfrac{3}{4}\\ \Rightarrow\left[{}\begin{matrix}x-\dfrac{2}{5}=\dfrac{3}{4}\\x-\dfrac{2}{5}=\dfrac{-3}{4}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{23}{20}\\x=\dfrac{-7}{20}\end{matrix}\right.\)
b.
\(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{2y}{6}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{2y}{6}=\dfrac{x+2y-z}{5+6-4}=\dfrac{14}{7}=2\\ \Rightarrow\left\{{}\begin{matrix}x=10\\y=6\\z=8\end{matrix}\right.\)
a.Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) => \(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có: \(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\) (1)
\(\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}=\dfrac{\left(bk+dk\right)^2}{\left(b+d\right)^2}=\dfrac{k^2\left(b+d\right)^2}{\left(b+d\right)^2}=k^2\)(2)
Từ (1) và (2) suy ra: \(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(a+c\right)^2}{\left(b+d\right)^2}\)
b.M = \(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{50^2}\right)\)
= \(\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}...\dfrac{2499}{2500}\)
= \(\dfrac{1.3.2.4.3.5...49.51}{2^2.3^2.4^2...50^2}\)
\(\dfrac{51}{2.50}=\dfrac{51}{100}\)
Lời giải:
a)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\)
\(\Rightarrow \left(\frac{a}{b}\right)^2=\left(\frac{b}{d}\right)^2=\frac{(a+c)^2}{(b+d)^2}(1)\)
Mặt khác, \(\frac{a}{b}=\frac{c}{d}\Rightarrow \frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a^2+c^2}{b^2+d^2}(2)\) (áp dụng tính chất dãy tỉ số bằng nhau)
Từ \((1),(2)\Rightarrow \frac{(a+c)^2}{(b+d)^2}=\frac{a^2+c^2}{b^2+d^2}\)
b) Vì \(1-\frac{1}{2^2};1-\frac{1}{3^2};...;1-\frac{1}{50^2}<1\) nên:
\(\left\{\begin{matrix} \left \{ 1-\frac{1}{2^2} \right \}=1-\frac{1}{2^2}\\ \left \{ 1-\frac{1}{3^2} \right \}=1-\frac{1}{3^2}\\ ....\\ \left \{ 1-\frac{1}{50^2} \right \}=1-\frac{1}{50^2}\end{matrix}\right.\)
\(\Rightarrow M=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)....\left(1-\frac{1}{50^2}\right)\)
\(\Leftrightarrow M=\frac{(2^2-1)(3^2-1)(4^2-1)....(50^2-1)}{(2.3....50)^2}\)
\(\Leftrightarrow M=\frac{[(2-1)(3-1)...(50-1)][(2+1)(3+1)...(50+1)]}{(2.3.4...50)^2}\)
\(\Leftrightarrow M=\frac{(2.3...49)(3.4.5...51)}{(2.3.4...50)^2}=\frac{(2.3.4...49)^2.50.51}{2.(2.3....49)^2.50^2}=\frac{50.51}{2.50^2}=\frac{51}{100}\)
Bài 1:
Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{x+y}{3+5}=\dfrac{16}{8}=2\)
\(\Rightarrow\left\{{}\begin{matrix}x=6\\y=10\end{matrix}\right.\)
Vậy x = 6, y = 10
Bài 2:
Ta có: \(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\)
\(\Rightarrow\left(a+5\right)\left(b-6\right)=\left(a-5\right)\left(b+6\right)\)
\(\Rightarrow ab-6a+5b-30=ab+6a-5b-30\)
\(\Rightarrow-6a+5b=6a-5b\)
\(\Rightarrow10b=12a\)
\(\Rightarrow6a=5b\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{5}{6}\)
\(\Rightarrowđpcm\)
B1 :
+ Theo bài ra :
\(\dfrac{x}{3}=\dfrac{y}{5}\left(1\right)\)và \(x+y=16\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{x+y}{3+5}=\dfrac{16}{8}=2\)
+ Do đó :
\(\dfrac{x}{3}=2\Rightarrow x=2.3=6\)
\(\dfrac{y}{5}=2\Rightarrow y=2.5=10\)
Vậy x = 6 ; y = 10
2.
\(\dfrac{a}{b}< \dfrac{c}{d}\Rightarrow ad< bc\) . Ta có : +,ad < bc
\(\Rightarrow\)ad+ab < bc +ab (Cùng thêm ab vào 2 vế)
\(\Rightarrow\)a(b+d) < b(a+c)
\(\Rightarrow\)\(\dfrac{a}{b}\)< \(\dfrac{a+c}{b+d}\)
+, ad < bc
\(\Rightarrow\)ad + cd < bc + cd ( Cùng thêm cd vào 2 vế)
\(\Rightarrow\)d(a+c) < c(b+d)
\(\Rightarrow\)\(\dfrac{a+c}{b+d}< \dfrac{c}{d}\) Vậy \(\dfrac{a}{b}< \dfrac{a+c}{b+d}< \dfrac{c}{d}\)
2.
ta có
\(\dfrac{a}{b}< \dfrac{c}{d}\Leftrightarrow\dfrac{ad}{bd}< \dfrac{bc}{bd}\Rightarrow ad< bc\)
xét
\(\dfrac{a}{b}=\dfrac{a\left(b+d\right)}{b\left(b+d\right)}=\dfrac{ab+ad}{b\left(b+d\right)}\)
\(\dfrac{a+c}{b+d}=\dfrac{b\left(a+c\right)}{b\left(b+d\right)}=\dfrac{ab+bc}{b\left(b+d\right)}\)
vì \(\dfrac{ab+ad}{b\left(b+d\right)}< \dfrac{ab+bc}{b\left(b+d\right)}\left(ad< bc\right)\)
\(\Rightarrow\dfrac{a}{b}< \dfrac{a+c}{b+d}\left(1\right)\)
xét
\(\dfrac{a+c}{b+d}=\dfrac{d\left(a+c\right)}{d\left(b+d\right)}=\dfrac{ad+cd}{d\left(b+d\right)}\)
\(\dfrac{c}{d}=\dfrac{c\left(b+d\right)}{d\left(b+d\right)}=\dfrac{bc+cd}{d\left(b+d\right)}\)
vì
\(\dfrac{ad+cd}{d\left(b+d\right)}< \dfrac{bc+cd}{d\left(b+d\right)}\left(ad< bc\right)\)
\(\Rightarrow\dfrac{a+c}{b+d}< \dfrac{c}{d}\left(2\right)\)
từ (1) và (2) => ĐPCM
1)\(\dfrac{x+1}{-12}=\dfrac{-3}{x+1}\)
\(\Rightarrow\left(x+1\right)^2=36\)
\(\Rightarrow\left[{}\begin{matrix}x+1=6\\x+1=-6\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-7\end{matrix}\right.\)
Vậy....
b)\(\left(\dfrac{1}{2}-2^2:\dfrac{4}{3}\right).\dfrac{6}{5}-7\)
\(=\left(\dfrac{1}{2}-4.\dfrac{3}{4}\right).\dfrac{6}{5}-7\)
\(=\left(\dfrac{1}{2}-3\right).\dfrac{6}{5}-7\)
\(=\dfrac{-5}{2}.\dfrac{6}{5}-7\)
\(=-3-7\)
\(=-10\)
Câu 1:
1/ Tìm x:(mk nghĩ là z)
\(\dfrac{x+1}{-12}=\dfrac{-3}{x+1}\Rightarrow\left(x+1\right)^2=\left(-3\right).\left(-12\right)=36\)
\(\Rightarrow x+1=6;x+1=-6\)
+) \(x+1=6\Rightarrow x=5\)
+) \(x+1=-6\Rightarrow x=-7\)
2/Tính:
\(\left(\dfrac{1}{2}-2^2:\dfrac{4}{3}\right).\dfrac{6}{5}-7=\left(\dfrac{1}{2}-\dfrac{4.3}{4}\right).\dfrac{6}{5}-7\)
\(=\left(\dfrac{1}{2}-3\right).\dfrac{6}{5}-7=\left(\dfrac{1}{2}.\dfrac{6}{5}\right)-\left(3.\dfrac{6}{5}\right)-7\)
\(=0,6-3,6-7=-10\)