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\(A=x^2-xy+\frac{y^2}{4}+\frac{3}{4}\left(y^2-4y+4\right)+2013\)
\(=\left(x-\frac{y}{2}\right)^2+\frac{3}{4}\left(y-2\right)^2+2013\ge2013\)
\(B\) đề thiếu
\(C\) đề sai, dấu của \(y^2\) là âm thì không tồn tại GTNN
\(P=-\left(x^2-2x+1\right)-\left(4y^2+4y+1\right)+7\)
\(=-\left(x-1\right)^2-\left(2y+1\right)^2+7\le7\)
\(2Q=-4x^2-20y^2+12xy+8x-6y+4\)
\(=-\left(4x^2+9y^2+4-12xy-8x+12y\right)-11\left(y^2-\frac{6}{11}y+\frac{36}{121}\right)+\frac{97}{11}\)
\(=-\left(2x-3y-2\right)^2-11\left(y-\frac{3}{11}\right)^2+\frac{97}{11}\le\frac{97}{11}\)
\(\Rightarrow Q\le\frac{97}{22}\)
a) Xem lại đề
b) x³ - 4x²y + 4xy² - 9x
= x(x² - 4xy + 4y² - 9)
= x[(x² - 4xy + 4y² - 3²]
= x[(x - 2y)² - 3²]
= x(x - 2y - 3)(x - 2y + 3)
c) x³ - y³ + x - y
= (x³ - y³) + (x - y)
= (x - y)(x² + xy + y²) + (x - y)
= (x - y)(x² + xy + y² + 1)
d) 4x² - 4xy + 2x - y + y²
= (4x² - 4xy + y²) + (2x - y)
= (2x - y)² + (2x - y)
= (2x - y)(2x - y + 1)
e) 9x² - 3x + 2y - 4y²
= (9x² - 4y²) - (3x - 2y)
= (3x - 2y)(3x + 2y) - (3x - 2y)
= (3x - 2y)(3x + 2y - 1)
f) 3x² - 6xy + 3y² - 5x + 5y
= (3x² - 6xy + 3y²) - (5x - 5y)
= 3(x² - 2xy + y²) - 5(x - y)
= 3(x - y)² - 5(x - y)
= (x - y)[(3(x - y) - 5]
= (x - y)(3x - 3y - 5)
Ta có:
D=2x2+3y2+4xy−8x−2y+18C=2x2+3y2+4xy−8x−2y+18
D=2(x2+2xy+y2)+y2−8x−2y+18C=2(x2+2xy+y2)+y2−8x−2y+18
D=2[(x+y)2−4(x+y)+4]+(y2+6y+9)+1C=2[(x+y)2−4(x+y)+4]+(y2+6y+9)+1
D=2(x+y−2)2+(y+3)2+1≥1C=2(x+y−2)2+(y+3)2+1≥1
Dấu "=" xảy ra ⇔x+y=2⇔x+y=2và y=−3y=−3
Hay x = 5 , y = -3
Đc chx bạn
a) \(A=5x\left(4x^2-2x+1\right)-2x\left(10x^2-5x-2\right)\)
\(A=20x^3-10x^2+5x-20x^3+10x^2+4x\)
\(A=9x\)
Thay x = 15 vào, ta có:
\(A=9.15=135\)
b) \(B=5x\left(x-4y\right)-4y\left(y-5x\right)\)
\(B=5x^2-20xy-4y^2+20xy\)
\(B=5x^2-4y\)
Thay \(x=-\frac{1}{5};y=-\frac{1}{2}\) vào, ta có:
\(B=5.\left(-\frac{1}{5}\right)^2-4.\left(-\frac{1}{2}\right)=\frac{11}{5}\)
c) \(C=6xy\left(xy-y^2\right)-8x^2\left(x-y^2\right)-5y^2\left(x^2-xy\right)\)
\(C=6x^2y^2-6xy^3-8x^3+8x^2y^2-5x^2y^2+5xy^3\)
\(C=9x^2y^2-xy^3-8x^3\)
Thay \(x=\frac{1}{2};y=2\) vào, ta có:
\(C=9.\left(\frac{1}{2}\right)^2.2^2-\frac{1}{2}.2^3-8.\left(\frac{1}{2}\right)^3=4\)
d) \(D=\left(3x+5\right)\left(2x-1\right)+\left(4x-1\right)\left(3x+2\right)\)
\(D=6x^2-3x+10x-5+12x^2+8x-3x-2\)
\(D=18x^2+12x-7\)
Ta có: \(\left|2\right|=\orbr{\begin{cases}x=-2\\x=2\end{cases}}\)
+) Với x = -2
\(D=18.\left(-2\right)^2+12.\left(-2\right)-7=41\)
+) Với x = 2
\(D=18.2^2+12.2-7=89\)
a/ \(A=20x^3-10x^2+5x-20x^3+10x^2+4x=9x\)
Thay x = 15 vào bt A ta có
A = 9 . 15 = 135
b/ \(B=5x^2-20xy-4y^2+2xy=5x^2-4y^2\)
Thay x = -1/5 ; y = - 1/2 vào bt B ta có
\(B=5.\dfrac{1}{25}-4.\dfrac{1}{4}=\dfrac{1}{5}-1=-\dfrac{4}{5}\)
c/ \(C=6x^2y^2-6xy^3-8x^3+8x^2y^2-5x^2y^2+5xy^3\)
\(=9x^2y^2-xy^3-8x^3\)
Thay x = 1/2 ; y = 2 vào bt C ta có
\(C=9.4.\dfrac{1}{4}-\dfrac{1}{2}.8-8.\dfrac{1}{8}=9-4-1=4\)
d/ \(D=6x^2+10x-3x-5+6x^2-3x+8x-2\)
\(=12x^2+12x-3\)
\(\left|x\right|=2\Rightarrow x=\pm2\)
Thay x = 2 vào bt D có
\(D=12.4+12.2-3=69\)
Thay x = - 2 vào bt D ta có
\(D=12.4-12.2-3=21\)
Bài 1:
a: ĐKXĐ: \(x+4\ne0\)
=>\(x\ne-4\)
b: ĐKXĐ: \(2x-1\ne0\)
=>\(2x\ne1\)
=>\(x\ne\dfrac{1}{2}\)
c: ĐKXĐ: \(x\left(y-3\right)\ne0\)
=>\(\left\{{}\begin{matrix}x\ne0\\y-3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\y\ne3\end{matrix}\right.\)
d: ĐKXĐ: \(x^2-4y^2\ne0\)
=>\(\left(x-2y\right)\left(x+2y\right)\ne0\)
=>\(x\ne\pm2y\)
e: ĐKXĐ: \(\left(5-x\right)\left(y+2\right)\ne0\)
=>\(\left\{{}\begin{matrix}x\ne5\\y\ne-2\end{matrix}\right.\)
Bài 2:
a: \(\dfrac{-12x^3y^2}{-20x^2y^2}=\dfrac{12x^3y^2}{20x^2y^2}=\dfrac{12x^3y^2:4x^2y^2}{20x^2y^2:4x^2y^2}=\dfrac{3x}{5}\)
b: \(\dfrac{x^2+xy-x-y}{x^2-xy-x+y}\)
\(=\dfrac{\left(x^2+xy\right)-\left(x+y\right)}{\left(x^2-xy\right)-\left(x-y\right)}\)
\(=\dfrac{x\left(x+y\right)-\left(x+y\right)}{x\left(x-y\right)-\left(x-y\right)}=\dfrac{\left(x+y\right)\left(x-1\right)}{\left(x-y\right)\left(x-1\right)}\)
\(=\dfrac{x+y}{x-y}\)
c: \(\dfrac{7x^2-7xy}{y^2-x^2}\)
\(=\dfrac{7x\left(x-y\right)}{\left(y-x\right)\left(y+x\right)}\)
\(=\dfrac{-7x\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}=\dfrac{-7x}{x+y}\)
d: \(\dfrac{7x^2+14x+7}{3x^2+3x}\)
\(=\dfrac{7\left(x^2+2x+1\right)}{3x\left(x+1\right)}\)
\(=\dfrac{7\left(x+1\right)^2}{3x\left(x+1\right)}=\dfrac{7\left(x+1\right)}{3x}\)
e: \(\dfrac{3y-2-3xy+2x}{1-3x-x^3+3x^2}\)
\(=\dfrac{3y-2-x\left(3y-2\right)}{1-3x+3x^2-x^3}\)
\(=\dfrac{\left(3y-2\right)\left(1-x\right)}{\left(1-x\right)^3}=\dfrac{3y-2}{\left(1-x\right)^2}\)
g: \(\dfrac{x^2+7x+12}{x^2+5x+6}\)
\(=\dfrac{\left(x+3\right)\left(x+4\right)}{\left(x+3\right)\left(x+2\right)}\)
\(=\dfrac{x+4}{x+2}\)
Bài 1:
a)
\(A=x^2+y^2-xy-3y+2016=(x^2-xy+\frac{y^2}{4})+(\frac{3y^2}{4}-3y+3)+2013\)
\(=(x-\frac{y}{2})^2+3(\frac{y}{2}-1)^2+2013\)
\(\geq 2013\)
Vậy GTNN của $A$ là $2013$. Giá trị này đạt được khi \(\left\{\begin{matrix} x-\frac{y}{2}=0\\ \frac{y}{2}-1=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} y=2\\ x=1\end{matrix}\right.\)
b)
\(B=2x^2+5y^2+4xy-6+5x-9\)
\(=5(y^2+\frac{4}{5}xy+\frac{4}{25}x^2)+\frac{6}{5}x^2+5x-15\)
\(=5(y+\frac{2}{5}x)^2+\frac{6}{5}(x^2+\frac{25}{6}x+\frac{25^2}{12^2})-\frac{485}{24}\)
\(=5(y+\frac{2}{5}x)^2+\frac{6}{5}(x+\frac{25}{12})^2-\frac{485}{24}\geq \frac{-485}{24}\)
Vậy GTNN của $B$ là $\frac{-485}{24}$
Giá trị này đạt được khi \(\left\{\begin{matrix} y+\frac{2}{5}x=0\\ x+\frac{25}{12}=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=-\frac{25}{12}\\ y=\frac{5}{6}\end{matrix}\right.\)
c)
\(C=x^2+xy+y^2-3x-3y+2018\)
\(=\frac{4x^2+4xy+4y^2-12x-12y+8072}{4}=\frac{(4x^2+4xy+y^2)+3y^2-12x-12y+8072}{4}\)
\(=\frac{(2x+y)^2-6(2x+y)+3y^2-6y+8072}{4}\)
\(=\frac{(2x+y)^2-6(2x+y)+9+3(y^2-2y+1)+8060}{4}=\frac{(2x+y-3)^2+3(y-1)^2+8060}{4}\)
\(\geq \frac{8060}{4}=2015\)
Vậy $C_{\min}=2015$. Giá trị đạt được khi \(\left\{\begin{matrix} 2x+y-3=0\\ y-1=0\end{matrix}\right.\Leftrightarrow x=y=1\)
Bài 2:
a)
\(-A=x^2+4y^2-2x+4y-5=(x^2-2x+1)+(4y^2+4y+1)-7\)
\(=(x-1)^2+(2y+1)^2-7\geq -7\)
\(\Rightarrow A\leq 7\)
Vậy GTLN của $A$ là $7$.
Giá trị này đạt được khi \(\left\{\begin{matrix} x-1=0\\ 2y+1=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=1\\ y=\frac{-1}{2}\end{matrix}\right.\)
b)
ĐKĐB \(\Leftrightarrow B+2x^2+10y^2-6xy-4x+3y-2=0\)
\(\Leftrightarrow 2x^2-2x(3y+2)+(10y^2+3y-2+B)=0\)
Coi đây là PT bậc 2 ẩn $x$. Vì dấu "=" tồn tại nên PT luôn có nghiệm
\(\Rightarrow \Delta'=(3y+2)^2-2(10y^2+3y-2+B)\geq 0\)
\(\Leftrightarrow B\leq \frac{-11y^2+6y+8}{2}=\frac{\frac{97}{11}-11(y-\frac{3}{11})^2}{2}\leq \frac{97}{22}\)
Vậy $B_{\max}=\frac{97}{22}$