Tìm x,y nguyên biết
a/ |x - 3| + |2y - 6| + 10 = \(\dfrac{30}{\left(y-3\right)^2+3}\)
b/ (2x + 6)2020 + 51 = \(\dfrac{102}{3\left|x+3\right|+2}\)
a, \(\text{[}\left(x-y\right)^3+3\left(x-y\right)\text{]}:\dfrac{1}{3}\left(x-y\right)\)
b, \(\left(8x^3-27y^3\right):\left(2x-3y\right)\)
c, \(\text{[}5\left(x+2y\right)^6-6\left(x+2y\right)^5\text{]}:2\left(x+2y\right)^4\)
a: \(=\left(x-y\right)^3:\dfrac{1}{3}\left(x-y\right)+3\left(x-y\right):\dfrac{1}{3}\left(x-y\right)\)
=3(x-y)^2+9
b: \(=\dfrac{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}{2x-3y}=4x^2+6xy+9y^2\)
c: \(=\dfrac{5\left(x+2y\right)^6}{2\left(x+2y\right)^4}-\dfrac{6\left(x+2y\right)^5}{2\left(x+2y\right)^4}=\dfrac{5}{2}\left(x+2y\right)^2-3\left(x+2y\right)\)
Bài 1: Tính:
a)\(\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}-\dfrac{2y^2}{y^2-x^2}\)
b)\(\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3}-\dfrac{x}{3x+9}\right)\)
Bài 2: Tìm x:
a)2x\(^3\)-50x=0 b)\(x^3+x^2+x+a\) chia hết cho x+1
Bài 3: Cho △MNP vuông tại N, biết MN = 6cm, NP = 8cm. đường cao NH, qua H kẻ HC⊥MN, HD⊥NP
a) Chứng minh HDNC là hình chữ nhật.
b) Tính CD
c) Tính diện tích △NMH
Bài 1:
\(a,=\dfrac{x^2+2xy+y^2-x^2+2xy-y^2+2y^2}{2\left(x-y\right)\left(x+y\right)}=\dfrac{2y\left(x+y\right)}{2\left(x-y\right)\left(x+y\right)}=\dfrac{y}{x-y}\\ b,Sửa:\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\\ =\dfrac{9+x^2-3x}{x\left(x-3\right)\left(x+3\right)}:\dfrac{3x-9-x^2}{3x\left(x+3\right)}=\dfrac{x^2+3x+9}{x\left(x-3\right)\left(x+3\right)}\cdot\dfrac{-3x\left(x+3\right)}{x^2-3x+9}\\ =\dfrac{-3}{x-3}\)
Bài 2:
\(a,\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\\ b,\Leftrightarrow x^3+x^2+x+a=\left(x+1\right)\cdot a\left(x\right)\\ \text{Thay }x=-1\Leftrightarrow-1+1-1+a=0\Leftrightarrow a=1\)
Giải hệ
a) \(\left\{{}\begin{matrix}x^2+y^2-2y-6+2\sqrt{2y+3}=0\\\left(x-y\right)\left(x^2+xy+y^2+3\right)=3\left(x^2+y^2\right)+2\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x^2y+2y+x=4xy\\\dfrac{1}{x^2}+\dfrac{1}{xy}+\dfrac{x}{y}=3\end{matrix}\right.\)
Tìm x, y biết :
\(\left|y+3\right|+5=\dfrac{10}{\left(2x-6\right)^2+2}\)
Ta có: \(\left|y+3\right|\ge0\Rightarrow\left|y+3\right|+5\ge5\)
\(\left(2x-6\right)^2\ge0\Rightarrow\left(2x-6\right)^2+2\ge2\)
\(\Rightarrow\dfrac{10}{\left(2x-6\right)^2+2}\le5\)
Để pt có nghiệm <=> \(\left[{}\begin{matrix}2x-6=0\\y+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\y=-3\end{matrix}\right.\)
Vậy x=3, y=-3
1, Rút gọn các phân thức sau :
a, \(\dfrac{5x}{10}\)
b, \(\dfrac{4xy}{2y}\) ( y # 0)
c, \(\dfrac{21x^2y^3}{6xy}\) ( xy # 0)
d, \(\dfrac{2x+2y}{4}\)
e, \(\dfrac{5x-5y}{3x-3y}\) ( x # y)
f, \(\dfrac{-15x\left(x-y\right)}{3\left(y-x\right)}\) ( x # y)
2, Rút gọn các phân thức sau :
a, \(\dfrac{x^2-16}{4x-x^2}\) ( x # 0, x # 4)
b, \(\dfrac{x^2+4x+3}{2x+6}\) ( x # -3)
c, \(\dfrac{15x\left(x+3\right)^3}{5y\left(x+y\right)^2}\) ( y + ( x+y) # 0)
d, \(\dfrac{5\left(x-y\right)-3\left(y-x\right)}{10\left(x-y\right)}\) ( x # y)
e, \(\dfrac{2x+2y+5x+5y}{2x+2y-5x-5y}\) (x # -y)
1)
a) \(\dfrac{5x}{10}=\dfrac{x}{2}\)
b) \(\dfrac{4xy}{2y}=2x\left(y\ne0\right)\)
c) \(\dfrac{21x^2y^3}{6xy}=\dfrac{7xy^2}{2}\left(xy\ne0\right)\)
d) \(\dfrac{2x+2y}{4}=\dfrac{2\left(x+y\right)}{4}=\dfrac{x+y}{2}\)
e) \(\dfrac{5x-5y}{3x-3y}=\dfrac{5\left(x-y\right)}{3\left(x-y\right)}=\dfrac{5}{3}\left(x\ne y\right)\)
f) \(\dfrac{-15x\left(x-y\right)}{3\left(y-x\right)}=-5x\dfrac{x-y}{y-x}=-5x\dfrac{x-y}{-\left(x-y\right)}\)
\(=-5x.\left(-1\right)=5x\left(x\ne y\right)\)
2)
a) Nhớ ghi ĐK vào nhá, lười quá :V\(\dfrac{x^2-16}{4x-x^2}=-\dfrac{\left(x-4\right)\left(x+4\right)}{x^2-4x}=\dfrac{\left(x-4\right)\left(x+4\right)}{x\left(x-4\right)}=\dfrac{x+4}{x}\)
b) \(\dfrac{x^2+4x+3}{2x+6}=\dfrac{x^2+3x+x+3}{2\left(x+3\right)}=\dfrac{x\left(x+3\right)+\left(x+3\right)}{2\left(x+3\right)}\)
\(=\dfrac{\left(x+3\right)\left(x+1\right)}{2\left(x+3\right)}=\dfrac{x+1}{2}\)
c) \(\dfrac{15x\left(x+3\right)^3}{5y\left(x+y\right)^2}=\dfrac{3x\left(x+3\right)^3}{y\left(x+y\right)^2}\) ( câu này có gì đó sai sai )
d) \(\dfrac{5\left(x-y\right)-3\left(y-x\right)}{10\left(x-y\right)}=\dfrac{5\left(x-y\right)+3\left(x-y\right)}{10\left(x-y\right)}\)
\(=\dfrac{8\left(x-y\right)}{10\left(x-y\right)}=\dfrac{8}{10}=\dfrac{4}{5}\)
e) \(\dfrac{2x+2y+5x+5y}{2x+2y-5x-5y}=\dfrac{2\left(x+y\right)+5\left(x+y\right)}{2\left(x+y\right)-5\left(x+y\right)}\)
\(=\dfrac{7\left(x+y\right)}{-3\left(x+y\right)}=-\dfrac{7}{3}\)
\(a.\left|x+2\right|+\left|x-1\right|=3-\left(y+2\right)^2\)
\(b.\left|x-5\right|+\left|1-x\right|=\dfrac{12}{\left|y+1\right|+3}\)
\(c.\left|y+3\right|+5=\dfrac{10}{\left(2x-6\right)^2+2}\)
\(d.\left|x-1\right|+\left|3-x\right|=\dfrac{6}{\left|y-3\right|+3}\)
a) Áp dụng bất đẳng thức \(\left|A\right|+\left|B\right|\ge\left|A+B\right|\) ta có :
\(\left|x+2\right|+\left|x-1\right|=\left|x+2\right|+\left|1-x\right|\)
\(\ge\left|x+2+1-x\right|=3\) (1)
Dấu "=" xảy ra \(\Leftrightarrow\left(x+2\right)\left(1-x\right)\ge0\)
\(\Leftrightarrow-2\le x\le1\)
+ \(\left(y+2\right)^2\ge0\forall y\)
\(\Rightarrow3-\left(y+2\right)^2\le3\) (2)
Dấu "=" xảy ra \(\Leftrightarrow\left(y+2\right)^2=0\Leftrightarrow y=-2\)
Từ (1) và (2) suy ra \(\left|x+2\right|+\left|x+1\right|=3-\left(y+2\right)^2=3\)
\(\Leftrightarrow\left\{{}\begin{matrix}-2\le x\le1\\y=-2\end{matrix}\right.\)
b) \(\left|x-5\right|+\left|1-x\right|\ge\left|x-5+1-x\right|=4\) (3)
Dấu "=" xảy ra \(\Leftrightarrow\left(x-5\right)\left(1-x\right)\ge0\)
\(\Leftrightarrow1\le x\le5\)
+ \(\left|y+1\right|\ge0\forall y\) \(\Rightarrow\left|y+1\right|+3\ge3\)
\(\Rightarrow\frac{12}{\left|y+1\right|+3}\le\frac{12}{3}=4\) (4)
Dấu "=" xảy ra \(\Leftrightarrow\left|y+1\right|=0\Leftrightarrow y=-1\)
Từ (3) và (4) suy ra \(\left|x-5\right|+\left|1-x\right|=\frac{12}{\left|y+1\right|+3}=4\)
\(\Leftrightarrow\left\{{}\begin{matrix}1\le x\le5\\y=-1\end{matrix}\right.\)
Câu c,d lm tương tự
Tìm x
\(a,3-x=x+1,8\)
\(b,2x-5=7x+35\)
\(c,2\left(x+10\right)=3\left(x-6\right)\)
\(d,8\left(x-\dfrac{3}{8}\right)+1=6\left(\dfrac{1}{6}+x\right)+x\)
\(e,\dfrac{2}{9}-3x=\dfrac{4}{3}-x\)
\(g,\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{3}{4}x-\dfrac{1}{2}\)
\(h,x-4=\dfrac{5}{6}\left(6-\dfrac{6}{5}x\right)\)
\(k,7x^2-11=6x^2-2\)
\(m,5\left(x+3.2^3\right)=10^2\)
\(n,\dfrac{4}{9}-(\dfrac{1}{6^2})=\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}\)
\(a,3-x=x+1,8\)
\(\Rightarrow-x-x=1,8-3\)
\(\Rightarrow-2x=-1,2\)
\(\Rightarrow x=0,6\)
\(b,2x-5=7x+35\)
\(\Rightarrow2x-7x=35+5\)
\(\Rightarrow-5x=40\)
\(\Rightarrow x=-8\)
\(c,2\left(x+10\right)=3\left(x-6\right)\)
\(\Rightarrow2x+20=3x-18\)
\(\Rightarrow2x-3x=-18-20\)
\(\Rightarrow-x=-38\)
\(\Rightarrow x=38\)
\(d,8\left(x-\dfrac{3}{8}\right)+1=6\left(\dfrac{1}{6}+x\right)+x\)
\(\Rightarrow8x-3+1=1+6x+x\)
\(\Rightarrow8x-3=7x\)
\(\Rightarrow8x-7x=3\)
\(\Rightarrow x=3\)
\(e,\dfrac{2}{9}-3x=\dfrac{4}{3}-x\)
\(\Rightarrow-3x+x=\dfrac{4}{3}-\dfrac{2}{9}\)
\(\Rightarrow-2x=\dfrac{10}{9}\)
\(\Rightarrow x=-\dfrac{5}{9}\)
\(g,\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{3}{4}x-\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{2}x-\dfrac{3}{4}x=-\dfrac{1}{2}-\dfrac{5}{6}\)
\(\Rightarrow-\dfrac{1}{4}x=-\dfrac{4}{3}\)
\(\Rightarrow x=\dfrac{16}{3}\)
\(h,x-4=\dfrac{5}{6}\left(6-\dfrac{6}{5}x\right)\)
\(\Rightarrow x-4=5-x\)
\(\Rightarrow x+x=5+4\)
\(\Rightarrow2x=9\)
\(\Rightarrow x=\dfrac{9}{2}\)
\(k,7x^2-11=6x^2-2\)
\(\Rightarrow7x^2-6x^2=-2+11\)
\(\Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
\(m,5\left(x+3\cdot2^3\right)=10^2\)
\(\Rightarrow5\left(x+24\right)=100\)
\(\Rightarrow x+24=20\)
\(\Rightarrow x=-4\)
\(n,\dfrac{4}{9}-\left(\dfrac{1}{6^2}\right)=\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{4}{9}-\dfrac{1}{36}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{5}{12}\)
\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2=0\)
\(\Rightarrow x-\dfrac{2}{3}=0\Rightarrow x=\dfrac{2}{3}\)
#\(Urushi\text{☕}\)
a: 3-x=x+1,8
=>-2x=-1,2
=>x=0,6
b: 2x-5=7x+35
=>-5x=40
=>x=-8
c: 2(x+10)=3(x-6)
=>3x-18=2x+20
=>x=38
d; 8(x-3/8)+1=6(1/6+x)+x
=>8x-3+1=1+6x+x
=>8x-2=7x+1
=>x=3
e: =>-3x+x=4/3-2/9
=>-2x=12/9-2/9=10/9
=>x=-5/9
g: =>3/4x-1/2x=5/6+1/2
=>1/4x=5/6+3/6=8/6=4/3
=>x=4/3*4=16/3
h: =>x-4=-x+5
=>2x=9
=>x=9/2
Tìm ĐKXĐ
\(\dfrac{1}{x^2+y^2};\dfrac{x^2y+2x}{x^2-2x+1};\dfrac{5x+y}{x^2+6x+10};\dfrac{x+y}{\left(x+3\right)^2+\left(y-2\right)^2}\)
a: ĐKXĐ: \(x^2+y^2\ne0\)
=>\(\left[{}\begin{matrix}x^2\ne0\\y^2\ne0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x\ne0\\y\ne0\end{matrix}\right.\)
b: ĐKXĐ: \(x^2-2x+1\ne0\)
=>\(\left(x-1\right)^2\ne0\)
=>\(x-1\ne0\)
=>\(x\ne1\)
c: ĐKXĐ: \(x^2+6x+10\ne0\)
=>\(x^2+6x+9+1\ne0\)
=>\(\left(x+3\right)^2+1\ne0\)(luôn đúng)
d:ĐKXĐ: \(\left(x+3\right)^2+\left(y-2\right)^2\ne0\)
=>\(\left[{}\begin{matrix}x+3\ne0\\y-2\ne0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x\ne-3\\y\ne2\end{matrix}\right.\)
GIẢI HPT
\(_{\left\{{}\begin{matrix}2\left(x+1\right)-\dfrac{y}{4}=10\\\dfrac{2x+3}{3-4x}=\dfrac{x-y+5}{6-2x+2y}\end{matrix}\right.\)