1+1=
2+2=
38929929*9
tìm x,y:a/9=b/8 biết a+b=34
tìm diệp
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Bài 1: Tìm y.
\(a)\dfrac{19}{20}-y=\dfrac{8}{5}-\dfrac{3}{4}\\ \dfrac{19}{20}-y=\dfrac{17}{20}\\ y=\dfrac{19}{20}-\dfrac{17}{20}\\ y=\dfrac{2}{20}\\ y=\dfrac{1}{10}\\ b)y:\dfrac{2}{3}=\dfrac{4}{9}\times3\\ y:\dfrac{2}{3}=\dfrac{4}{3}\\ y=\dfrac{4}{3}\times\dfrac{2}{3}\\ y=\dfrac{8}{9}\)
Bài 2: Giải:
Chiều cao của mảnh vườn đó là:
\(\dfrac{2}{3}-\dfrac{1}{4}=\dfrac{5}{12}\left(km\right)\)
Diện tích mảnh vườn đó là:
\(\dfrac{2}{3}\times\dfrac{5}{12}=\dfrac{5}{18}\left(km^2\right)\)
Đáp số: \(\dfrac{5}{18}km^2.\)
Bài 1:
a) \(\dfrac{19}{20}-y=\dfrac{8}{5}-\dfrac{3}{4}\)
\(\dfrac{19}{20}-y=\dfrac{17}{20}\)
\(y=\dfrac{19}{20}-\dfrac{17}{20}\)
\(y=\dfrac{2}{20}=\dfrac{1}{10}\)
b) \(y:\dfrac{2}{3}=\dfrac{4}{9}\times3\)
\(y:\dfrac{2}{3}=\dfrac{4}{3}\)
\(y=\dfrac{4}{3}\times\dfrac{2}{3}\)
\(y=\dfrac{8}{9}\)
Bài 2:
Giải:
Chiều cao của hình bình hành: \(\dfrac{2}{3}\times\dfrac{1}{4}=\dfrac{1}{6}\left(km\right)\)
Diện tích mảnh vườn: \(\dfrac{1}{6}\times\dfrac{2}{3}=\dfrac{1}{9}\left(km^2\right)\)
Đáp số: \(\dfrac{1}{9}km^2\)
Mẫu số to quá nên ko nghĩ ra cách giải đẹp mắt:
Dự đoán dấu "=" xảy ra tại \(a=b=c=1\), ta cần c/m: \(A\le\dfrac{3}{16}\)
Do \(\sum\dfrac{a+1}{a^2+1+10a+20}\le\sum\dfrac{a+1}{2a+10a+20}=\sum\dfrac{a+1}{12a+20}\)
Nên ta chỉ cần chứng minh: \(\sum\dfrac{a+1}{3a+5}\le\dfrac{3}{4}\Leftrightarrow\sum\left(\dfrac{3a+3}{3a+5}-1\right)\le\dfrac{9}{4}-3\)
\(\Leftrightarrow\sum\dfrac{1}{3a+5}\ge\dfrac{3}{8}\Leftrightarrow\dfrac{3\left(ab+bc+ca\right)+10\left(a+b+c\right)+25}{\left(3a+5\right)\left(3b+5\right)\left(3c+5\right)}\ge\dfrac{1}{8}\) (quy đồng)
\(\Leftrightarrow\dfrac{4\left(a+b+c\right)+3\left(ab+bc+ca+2\left(a+b+c\right)\right)+25}{27abc+45\left(ab+bc+ca+2\left(a+b+c\right)\right)-15\left(a+b+c\right)+125}\ge\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{4\left(a+b+c\right)+52}{27abc-15\left(a+b+c\right)+530}\ge\dfrac{1}{8}\)
\(\Leftrightarrow47\left(a+b+c\right)\ge27abc+114\)
Điều này đúng do:
\(9=2\left(a+b+c\right)+ab+bc+ca\le2\left(a+b+c\right)+\dfrac{1}{3}\left(a+b+c\right)^2\)
\(\Rightarrow\left(a+b+c-3\right)\left(a+b+c+9\right)\ge0\)
\(\Rightarrow a+b+c\ge3\)
Và: \(9=a+b+c+a+b+c+ab+bc+ca\ge9\sqrt[9]{a^4b^4c^4}\)
\(\Rightarrow abc\le1\)
\(\Rightarrow\left\{{}\begin{matrix}47\left(a+b+c\right)\ge141\\27abc+114\le27+114=141\end{matrix}\right.\) (đpcm)
Bài 2:
a: \(\dfrac{4\cdot7}{9\cdot32}=\dfrac{4}{32}\cdot\dfrac{7}{9}=\dfrac{1}{8}\cdot\dfrac{7}{9}=\dfrac{7}{72}\)
b: \(\dfrac{3\cdot21}{14\cdot15}=\dfrac{63}{210}=\dfrac{3}{10}\)
c: \(\dfrac{9\cdot6-9\cdot3}{18}=\dfrac{9\cdot3}{18}=\dfrac{3}{2}\)
d: \(\dfrac{17\cdot15-17}{3-20}=\dfrac{17\cdot14}{-17}=-14\)
e: \(=\dfrac{26\cdot5}{26\cdot35}=\dfrac{5}{35}=\dfrac{1}{7}\)
f: \(=\dfrac{49\left(1+7\right)}{49}=8\)
\(8x-75=5x+21\)
\(8x-5x=75+21\)
\(3x=96\)
\(x=32\)
Vậy \(x=32.\)
\(9x+25=-\left(2x-58\right)\)
\(9x+25=-2x+58\)
\(9x+2x=-25+58\)
\(11x=33\)
\(x=3\)
Vậy \(x=3.\)
\(15-\left|2x-1\right|=-8\)
\(\left|2x-1\right|=23\)
\(\Rightarrow\orbr{\begin{cases}2x-1=23\\2x-1=-23\end{cases}\Rightarrow}\orbr{\begin{cases}2x=24\\2x=-22\end{cases}}\Rightarrow\orbr{\begin{cases}x=12\\x=-11\end{cases}}\)
Vậy \(x\in\left\{12;-11\right\}\)
8x-5x=75+21
3x=96
x=32
9x+25=-2x+58
9x+2x=58-25
11x=33
x=3
\(a.\left(x^2+4x+4\right)+\left(x^2-6x+9\right)=2x^2+14x\)
\(x^2+4x+4+x^2-6x+9-2x^2-14x=0\)
\(-18x+13=0\)
\(x=\dfrac{13}{18}\)
Vậy \(S=\left\{\dfrac{13}{18}\right\}\)
\(b.\left(x-1\right)^3-125=0\)
\(\left(x-1\right)^3=125\)
\(x-1=5\)
\(x=6\)
Vậy \(S=\left\{6\right\}\)
\(c.\left(x-1\right)^2+\left(y +2\right)^2=0\)
\(Do\left(x-1\right)^2\ge0\forall x;\left(y+2\right)^2\ge0\forall y\)
\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0\forall x,y\)
Mà \(\left(x-1\right)^2+\left(y+2\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
Vậy \(S=\left\{1;-2\right\}\)
\(d.x^2-4x+4+x^2-2xy+y^2=0\)
\(\left(x-2\right)^2+\left(x-y\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-2\right)^2=0\\\left(x-y\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-y=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)
Vậy \(S=\left\{2;2\right\}\)
Bài 1:
b) \(2x+6⋮x-3\)
\(\Leftrightarrow2\left(x-3\right)+12⋮x-3\)
Mà \(2\left(x-3\right)⋮x-3\)
\(\Rightarrow12⋮x-3\)
làm nốt
d) \(x-1⋮2x+1\)
\(\Leftrightarrow2x-2⋮2x+1\)
\(\Leftrightarrow2x+1-3⋮2x+1\)
Mà \(2x+1⋮2x+1\)
\(\Rightarrow3⋮2x+1\)
Làm nốt
tìm a,b chứ bạn
Theo tính chất dãy tỉ số bằng nhau
\(\frac{a}{9}=\frac{b}{8}=\frac{a+b}{9+8}=\frac{34}{17}=2\)
\(\hept{\begin{cases}a=9.2=18\\b=8.2=16\end{cases}}\)