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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. \(\left(20x^4y-25x^2y^2-3x^2y\right):5x^2y\)
\(=4x^2-5y-\frac{3}{5}\)
b. \(\left(15xy^2+17xy^3+18y^2\right):6y^2\)
\(=\frac{5}{2}x+\frac{17}{6}xy+3\)
c. \(\left[3\left(x-y\right)^4+2\left(x-y\right)^3-5\left(x-y\right)^2\right]:\left(y-x\right)^2\)
\(=\left[3\left(x-y\right)^4+2\left(x-y\right)^3-5\left(x-y\right)^2\right]:\left(x-y\right)^2\)
\(=3\left(x-y\right)^2+2\left(x-y\right)-5\)
d. \(\left(x^2-2xy+y^2\right):\left(y-x\right)\)
\(=\left(x-y\right)^2:\left(y-x\right)\)
\(=\left(y-x\right)^2:\left(y-x\right)\)
\(=y-x\)
- Chỉ có vận dụng linh hoạt các phương pháp đã học thôi, chẳng có mẹo nào đâu :>>
Lời giải:
Theo bài ra ta có:
\(5x+2y=10\Leftrightarrow y=\frac{10-5x}{2}\)
Thay vào biểu thức $M$
\(M=3x.\frac{10-5x}{2}-x^2-\left(\frac{10-5x}{2}\right)^2\)
\(\Leftrightarrow 4M=-59x^2+160x-100\)
\(\Leftrightarrow 4M=\frac{500}{59}-59(x-\frac{80}{59})^2\)
Vì \((x-\frac{80}{59})^2\geq 0\Rightarrow 4M\leq \frac{500}{59}\Leftrightarrow M\leq \frac{125}{59}\)
Vậy \(M_{\max}=\frac{125}{59}\)
Dấu bằng xảy ra khi \(x=\frac{80}{59}; y=\frac{95}{59}\)
Nguyễn NamRibi Nkok NgokTrần Quốc LộcAnh TriêtQuang Ho SiPhạm Hoàng GiangThien Tu BorumThảo Phương
Nguyễn Thanh HằngTrương Hồng HạnhAkai HarumaHàn Vũ
Hoàng Thị Ngọc AnhNguyễn Huy ThắngAn Nguyễn BáPhương AnĐoàn Đức HiếuVõ Đông Anh TuấnNguyễn Phương Trâm
\(A.2\left(2x+x^2\right)-x^2\left(x+2\right)+\left(x^3-4x+3\right)\)
\(=4x+2x^2-x^3-2x^2+x^3-4x+3\)
\(=3\)
\(\Rightarrow A:\) đúng.
\(B.x\left(x^2+x+1\right)-x^2\left(x+1\right)-x+5\)
\(=x^3+x^2+x-x^2\left(x+1\right)-x-1+6\)
\(=x^3+x\left(x+1\right)-x^2\left(x+1\right)-\left(x+1\right)+6\)
\(=x^3+\left(x+1\right)\left(x-x^2-1\right)+6\)
\(=x^3+x^2+x-x^3-x^2-x-1+6\)
\(=5\)
\(\Rightarrow B:\) đúng.
\(C.3x\left(x-2\right)-5x\left(x-1\right)-8\left(x^2-3\right)\)
\(=3x^2-6x-5x^2+5x-8x^2+24\)
\(=-10x^2-x+24\)
\(\Rightarrow C:sai.\)
\(D.2y\left(y^2+y+1\right)-2y^2\left(y+1\right)-2\left(y+10\right)\)
\(=2y^3+2y+2y-2y^3-2y^2-2y-20\)
\(=-2y^2+2y-20\)
\(\Rightarrow D:sai.\)
a) \(-2x\left(10x-3\right)+5x\left(4x+1\right)=25\)
\(-20x^2+6x+20x^2+5x=25\)
\(\Rightarrow6x+5x=25\)
\(\Rightarrow11x=25\)
\(\Rightarrow x=\dfrac{25}{11}\)
b) \(y\left(5-2y\right)+2y\left(y-1\right)=15\)
\(5y-2y^2+2y^2-2y=15\)
\(\Rightarrow5y-2y=15\)
\(\Rightarrow3y=15\)
\(\Rightarrow y=5\)
c)\(x\left(x+1\right)-\left(x+1\right)=35\)
\(\Rightarrow\left(x-1\right)\left(x+1\right)=35\)
\(\Rightarrow x^2-1=35\)
\(\Rightarrow x^2=36\)
\(\Rightarrow x=6;x=-6\)
d)\(x\left(x^2+x+1\right)-x^2\left(x+1\right)=0\)
\(x^3+x^2+x-x^3+x=0\)
\(\Rightarrow x^2+2x=0\)
\(\Rightarrow x\left(x+2\right)=0\)
\(\Rightarrow x=0;x=0-2=-2\)
Vậy \(x=0;x=-2\)
Câu 1:
a: \(C=a^2+b^2=\left(a+b\right)^2-2ab=23^2-2\cdot132=265\)
b: \(D=x^3+y^3+3xy\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+3xy\)
\(=1-3xy+3xy=1\)
