Rút gọn phân thức
\(\frac{14xy^5\left(2x-3y\right)}{21x^2y\left(2x-3y\right)^2}\)
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Bài 1:
a) \(x^2-5x+1=0\)
\(\Leftrightarrow\left(x^2-5x+\frac{25}{4}\right)-\frac{21}{4}=0\)
\(\Leftrightarrow\left(x-\frac{5}{2}\right)^2-\frac{\left(\sqrt{21}\right)^2}{2^2}=0\)
\(\Leftrightarrow\left(x-\frac{5+\sqrt{21}}{2}\right)\left(x+\frac{\sqrt{21}-5}{2}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{5+\sqrt{21}}{2}=0\\x+\frac{\sqrt{21}-5}{2}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5+\sqrt{21}}{2}\\x=\frac{5-\sqrt{21}}{2}\end{cases}}\)
b) \(3x^2-12x-1=0\)
\(\Leftrightarrow3\left(x^2-4x+4\right)-13=0\)
\(\Leftrightarrow\left(x-2\right)^2-\left(\sqrt{\frac{13}{3}}\right)^2=0\)
\(\Leftrightarrow\left(x-2-\sqrt{\frac{13}{3}}\right)\left(x-2+\sqrt{\frac{13}{3}}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=2+\sqrt{\frac{13}{3}}\\x=2-\sqrt{\frac{13}{3}}\end{cases}}\)
Bài 2:
a) \(A=\frac{1}{4}x^2-x+1=\left(\frac{1}{2}x-1\right)^2\ge0\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left(\frac{1}{2}x-1\right)^2=0\Rightarrow\frac{1}{2}x=1\Rightarrow x=2\)
Vậy Min(A) = 0 khi x = 2
b) \(B=3x^2-4x-2=3\left(x^2-\frac{4}{3}x+\frac{4}{9}\right)-\frac{10}{3}=3\left(x-\frac{2}{3}\right)^2-\frac{10}{3}\ge-\frac{10}{3}\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(3\left(x-\frac{2}{3}\right)^2=0\Rightarrow x=\frac{2}{3}\)
Vậy \(Min\left(B\right)=-\frac{10}{3}\Leftrightarrow x=\frac{2}{3}\)
Pt \(\Leftrightarrow\left(4x-1\right)\sqrt{x^3+1}=2\left(x^3+x\right)+1\)
Đặt \(\sqrt{x^3+1}=i\)
Ta có pt : \(\left(4x-1\right)i=2i^2+1\)
\(\Leftrightarrow2i^2+\left(4x-1\right)i+2x-1=0\Leftrightarrow\orbr{\begin{cases}i=\frac{1}{2}\\i=2x-1\end{cases}}\)
Tới đây tự blabla tiếp:))
a) \(xy+3x+y=8\)
\(\Leftrightarrow\left(xy+3x\right)+\left(y+3\right)=11\)
\(\Leftrightarrow x\left(y+3\right)+\left(y+3\right)=11\)
\(\Leftrightarrow\left(x+1\right)\left(y+3\right)=11=1.11=\left(-1\right).\left(-11\right)\)
Ta xét các TH sau:
+ \(\hept{\begin{cases}x+1=1\\y+3=11\end{cases}}\Rightarrow\hept{\begin{cases}x=0\\y=8\end{cases}}\)
+ \(\hept{\begin{cases}x+1=11\\y+3=1\end{cases}}\Rightarrow\hept{\begin{cases}x=10\\y=-2\end{cases}}\)
+ \(\hept{\begin{cases}x+1=-1\\y+3=-11\end{cases}}\Rightarrow\hept{\begin{cases}x=-2\\y=-14\end{cases}}\)
+ \(\hept{\begin{cases}x+1=-11\\y+3=-1\end{cases}}\Rightarrow\hept{\begin{cases}x=-12\\y=-4\end{cases}}\)
Vậy ta có 4 cặp số (x;y) thỏa mãn: (0;8) ; (10;-2) ; (-2;-14) ; (-12;-4)
a. xy + 3x + y = 8
=> x ( y + 3 ) + ( y + 3 ) = 8 + 3 = 11
=> ( x + 1 ) ( y + 3 ) = 11
x + 1 | y + 3 | x | y |
11 | 1 | 10 | - 2 |
1 | 11 | 0 | 8 |
- 11 | - 1 | - 12 | - 4 |
- 1 | - 11 | - 2 | - 14 |
Vậy các cặp ( x ; y ) thỏa mãn đề bài là ( 10 ; - 2 ) ; ( 0 ; 8 ) ; ( - 12 ; - 4 ) ; ( - 2 ; - 14 )
b. Không rõ đề
thay xyz=2018 vào M ta có
\(M=\frac{xyz\cdot x}{xy+xyz\cdot x+xyz}+\frac{y}{yz+y+xyz}+\frac{z}{xz+z+1}\)
\(=\frac{x^2yz}{xy+x^2yz+xyz}+\frac{y}{yz+y+xyz}+\frac{z}{xz+x+1}\)
\(=\frac{x^2yz}{xy\left(1+xz+y\right)}+\frac{y}{y\left(z+1+xz\right)}+\frac{z}{xz+x+1}\)
\(=\frac{xz}{1+xz+y}+\frac{1}{z+1+xz}+\frac{z}{xz+1+xz}=\frac{xz+1+z}{z+1+xz}=1\)
Vậy M=1 với xyz=2018
Em chỉ làm đại thôi ạ, có gì sai mong chị bảo vì năm nay em mới lên lớp 7 :vv
\(M=\frac{2018x}{xy+2018x+2018}+\frac{y}{yz+y+2018}+\frac{z}{xz+z+1}\)
\(=\frac{2018x}{xy+2018x+2018}+\frac{xy}{xyz+xy+2018x}+\frac{xyz}{xyxz+xyz+xy}\)
\(=\frac{2018x}{xy+2018x+2018}+\frac{xy}{2018+xy+2018x}+\frac{2018}{xy+2018+2018x}\)
\(=\frac{2018x+xy+2018}{xy+2018x+2018}=1\)
Vậy M = 1.
a) Ta có : \(A=\frac{3x+5}{x+4}=\frac{3x+12-7}{x+4}=\frac{3\left(x+4\right)-7}{x+4}=3-\frac{7}{x+4}\)
Vì \(3\inℤ\Rightarrow\frac{-7}{x+4}\inℤ\Rightarrow-7⋮x+4\Rightarrow x+4\inƯ\left(-7\right)\)
=> \(x+4\in\left\{1;-1;-7;7\right\}\Rightarrow x\in\left\{-3;-5;-11;7\right\}\)
b) Ta có B = \(\frac{10x^2-7x-5}{2x-3}=\frac{10x^2-15x+8x-12+7}{2x-3}=\frac{5x\left(2x-3\right)+4\left(2x-3\right)+7}{2x-3}\)
\(=\frac{\left(5x+4\right)\left(2x-3\right)+7}{2x-3}=5x+4+\frac{7}{2x-3}\)
Vì \(\hept{\begin{cases}5x\inℤ\\4\inℤ\end{cases}\Rightarrow\frac{7}{2x-3}\inℤ\Rightarrow7⋮2x-3\Rightarrow2x-3\inƯ\left(7\right)\Rightarrow2x-3\in\left\{1;7;-1;-7\right\}}\)
=> \(x\in\left\{2;5;1;-2\right\}\)
a) Để A có nghĩa <=> \(\hept{\begin{cases}2x+10\ne0\\x\ne0\\2x\left(x+5\right)\ne0\end{cases}}\) <=> \(\hept{\begin{cases}x\ne-5\\x\ne0\\x\ne0;x\ne-5\end{cases}}\) <=> \(\hept{\begin{cases}x\ne-5\\x\ne0\end{cases}}\)
b) A = \(\frac{x^2+2x}{2x+10}+\frac{x-5}{x}+\frac{50-5x}{2x\left(x+5\right)}\)
A = \(\frac{x\left(x^2+2x\right)+2\left(x-5\right)\left(x+5\right)+50-5x}{2x\left(x+5\right)}\)
A = \(\frac{x^3+2x^2+2\left(x^2-25\right)+50-5x}{2x\left(x+5\right)}\)
A = \(\frac{x^3+2x^2+2x^2-50+50-5x}{2x\left(x+5\right)}\)
A = \(\frac{x^3+4x^2-5x}{2x\left(x+5\right)}\)
A = \(\frac{x\left(x^2+4x-5\right)}{2x\left(x+5\right)}\)
A = \(\frac{x^2+5x-x-5}{2\left(x+5\right)}\)
A = \(\frac{\left(x-1\right)\left(x+5\right)}{2\left(x+5\right)}=\frac{x-1}{2}\)
Bài làm :
\(\frac{14xy^5\left(2x-3y\right)}{21x^2y\left(2x-3y\right)^2}\)
\(=\frac{y^4}{\frac{3}{2}x\left(2x-3y\right)}\)
Chúc bạn học tốt !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!