Tìm GTNN
a)\(A=25x^2+3y^2-10x+11\)
b)\(B=\left(x-3\right)^2+\left(x-11\right)^2\)
c)\(C=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)\)
Tìm GTNN của biểu thức:
a) A=\(25x^2+3y^2-10x+11\)
b) B=\(\left(x-3\right)^2+\left(x-11\right)^2\)
c) C=\(\left(x+1\right).\left(x-2\right).\left(x-3\right).\left(x-6\right)\)
a) \(A=25x^2+3y^2-10x+11\)
\(A=\left(5x-1\right)^2+3y^2+11\ge11\forall x;y\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{1}{5}\\y=0\end{matrix}\right.\)
b) \(B=\left(x-3\right)^2+\left(x-11\right)^2\)
\(B=2\left(x^2-14x+65\right)\)
\(B=2\left[\left(x-7\right)^2+16\right]\)
\(B=2\left(x-7\right)^2+32\ge32\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=7\)
c) \(C=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)\)
\(C=\left(x^2-5x-6\right)\left(x^2-5x+6\right)\)
Đặt \(x^2-5x-6=a\)
\(C=a\left(a+12\right)\)
\(C=a^2+12a+36-36\)
\(C=\left(a+6\right)^2-36\ge-36\)
Dấu "=" xảy ra \(\Leftrightarrow a=-6\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
\(C=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)\\ C=\left(x+1\right)\left(x-6\right)\left(x-2\right)\left(x-3\right)\\ C=\left(x^2-5x-6\right)\left(x^2-5x+6\right)\\ C=\left(x^2-5x\right)^2-6^2\\ C=\left(x^2-5x\right)^2-36\)
Ta có:
\(\left(x^2-5x\right)^2\ge0\\ \Rightarrow C=\left(x^2-5x\right)^2-36\ge-36\)
Dấu "=" xảy ra khi và chỉ khi:
(x2 - 5x)2 = 0 => x2 - 5x = 0 => x(x - 5) = 0
=> x = 5 hoặc x = 0
Vậy MinC = -36 <=> x = 5; x = 0
Nhờ mn giúp mik với ạ
Tìm GTNN
A= \(\left(x-3y\right)^2+\left(2x-1\right)^4\)
B= \(\left|x-2\right|+\left|3x-2y\right|-4\)
C= \(\dfrac{-4}{\left|x+1\right|\left|y-3\right|+2}\)
D=\(\left|x-5\right|+\left|x-1\right|+7\)
Tìm x biết:
a) \(\left|x+2\dfrac{1}{2}\right|=\left|3x+1\right|\)
b) \(\left|2x-6\right|+\left|x+3\right|=8\)
c) \(2.\left|x+2\right|+\left|4-x\right|=11\)
\(c,\Rightarrow\left[{}\begin{matrix}-2\left(x+2\right)+\left(4-x\right)=11\left(x< -2\right)\\2\left(x+2\right)+\left(4-x\right)=11\left(-2\le x\le4\right)\\2\left(x+2\right)+\left(x-4\right)=11\left(x>4\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{11}{3}\left(tm\right)\\x=3\left(tm\right)\\x=\dfrac{11}{3}\left(ktm\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{11}{3}\end{matrix}\right.\)
\(a,\Rightarrow\left[{}\begin{matrix}x+\dfrac{5}{2}=3x+1\\x+\dfrac{5}{2}=-3x-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{7}{8}\end{matrix}\right.\)
\(b,\Rightarrow\left[{}\begin{matrix}6-2x-x-3=8\left(x\le-3\right)\\6-2x+x+3=8\left(-3\le x\le3\right)\\2x-6+x+3=8\left(x>3\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{-5}{3}\left(ktm\right)\\x=1\left(tm\right)\\x=\dfrac{11}{3}\left(tm\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{11}{3}\end{matrix}\right.\)
Thu gọn biểu thức
a) \(C=\frac{7}{9}x^3y^2\left(\frac{6}{11}axy^3\right)+\left(-5bx^2y^4\right)\left(\frac{-1}{2}axz\right)+ax\left(x^2y\right)^3\)
b)\(D=\frac{\left(3x^4y^4\right)^2\left(\frac{6}{11}x^3y\right)\left(8x^{n-7}\right)\left(-2x^{7-n}\right)}{15x^3y^2\left(0,4ax^2y^2z^2\right)^2}\)(với axyz khác 0)
\(C=\frac{7}{9}x^3y^2\left(\frac{6}{11}axy^3\right)+\left(-5bx^2y^4\right)\left(\frac{-1}{2}axz\right)+ax\left(x^2y\right)^3\)
\(\Rightarrow C=\frac{42}{9}ax^4y^5+\frac{5}{2}abx^3y^4z+ax\left(x^6y^3\right)\)
\(\Rightarrow C=\frac{42}{9}ax^4y^5+\frac{5}{2}abx^3y^4z+ax^7y^3\)
\(D=\frac{\left(3x^4y^4\right)^2\left(\frac{6}{11}x^3y\right)\left(8x^{n-7}\right)\left(-2x^{7-n}\right)}{15x^3y^2\left(0,4ax^2y^2z^2\right)^2}\)
\(D=\frac{\left[3.\frac{6}{11}.8.\left(-2\right)\right]\left(x^8x^3x^{n-7}x^{7-n}\right)\left(y^8y\right)}{15.0,4.\left(x^3x^4\right)\left(y^2y^4\right)z^4a}\)
\(D=\frac{\frac{-188}{11}x^{24}y^9}{6x^7y^6z^4a}\)
Làm tiếp bài của Song Ngư (๖ۣۜO๖ۣۜX๖ۣۜA)
\(D=\frac{\frac{-188}{11}x^{17}y^3}{6z^4a}\)
Thực hiện phép tính:
a) \(\dfrac{2}{5}xy\left(x^2y-5x+10y\right)\)
b) \(\left(x^2-1\right)\left(x^2+2x+y\right)\)
c) \(\left(x+3y\right)^2\)
d) \(\left(4x-y\right)^3\)
e) \(\left(x^2-2y\right)\left(x^2+2y\right)\)
g) \(18x^4y^2z:10x^4y\)
h) \(\left(x^3y^3+\dfrac{1}{2}x^2y^3-x^3y^2\right):\dfrac{1}{3}x^2y^2\)
i) \(\left(6x^3-7x^2-x+2\right):\left(2x+1\right)\)
k) \(\dfrac{5x-1}{3x^2y}+\dfrac{x+1}{3x^2y}\)
l) \(\dfrac{3x+1}{x^2-3x+1}+\dfrac{x^2-6x}{x^2-3x+1}\)
m) \(\dfrac{2x+3}{10x-4}+\dfrac{5-3x}{4-10x}\)
n) \(\dfrac{x}{x^2+2x+1}+\dfrac{3}{5x^2-5}\)
o) \(\dfrac{x^2+2}{x^3-1}+\dfrac{2}{x^2+x+1}+\dfrac{1}{1-x}\)
p) \(\dfrac{4x+2}{15x^3y}\dfrac{5y-3}{9x^2y}+\dfrac{x+1}{5xy^3}\)
q) \(\dfrac{2x-7}{10x-4}-\dfrac{3x+5}{4-10x}\)
r) \(\dfrac{3}{2x+6}-\dfrac{x-6}{2x^2+6x}\)
x) \(\dfrac{4y^2}{11x^4}.\left(-\dfrac{3x^2}{8y}\right)\)
y) \(\dfrac{x^2-4}{3x+12}.\dfrac{x+4}{2x-4}\)
z) \(\left(x^2-25\right):\dfrac{2x+10}{3x-7}\)
t) \(\left(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}\right):\dfrac{4x}{10x-5}\)
w) \(\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
c: \(=x^2+6xy+9y^2\)
e: \(=x^4-4y^2\)
1)rút gọn
a) (x+5)(\(x^2\) - 5x + 25) - \(\left(x+3\right)^3\) + (x-2)(\(x^2\) + 2x + 4) - \(\left(x-1\right)^3\)
b)\(\left(x+3y\right)^3+\left(3x+2y\right)\left(9x^2-6xy+4y^2\right)-\left(2y-3x\right)^3\)
c)\(\left(3x+y\right)^3-\left(5x-y\right)\left(25x^2+5xy+y^2\right)+\left(x+2y\right)^3\)
2)tìm x,biết
a)\(\left(x-1\right)^2-\left(x-2\right)\left(x+3\right)+\left(x+2\right)^3=\left(x-3\right)\left(x^2+3x+9\right)+6x\left(x+2\right)\)
Cảm ơn các bạn ^^
chứng minh rằng các biểu thức sau không phụ thuộc vào x:
a. \(A=\left(3x+7\right)\left(2x+3\right)-\left(3x-5\right)\left(2x+11\right)\)
b. \(B=\left(x^2-2\right)\left(x^2+x-1\right)-x\left(x^3+x^2-3x-2\right)\)
c. \(C=x\left(x^3+x^2-3x-2\right)-\left(x^2-2\right)\left(x^2+x-1\right)\)
Chào các bạn, hôm nay mình có một số bài toán cần các bạn giúp mình giải chúng:
1) Cho \(a+b+c=0\). Chứng minh:
a) \(\left(ab+bc+ac\right)^2=a^2b^2+b^2c^2+a^2c^2\)
b) \(a^4+b^4+c^4=2\left(ab+bc+ac\right)^2\)
2) Tìm GTNN của biểu thức:
a) \(A=25x^2+3y^2-10x+11\)
b) \(B=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)\)
Mong các bạn sẽ giúp mình.
1)
a) \(\left(ab+bc+ca\right)^2=a^2b^2+b^2c^2+c^2a^2+2\left(ab^2c+a^2bc+abc^2\right)\)\(=a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=a^2b^2+b^2c^2+c^2a^2\)(vì a+b+c=0)
b) \(a+b+c=0\Rightarrow a^2+b^2+c^2=-2\left(ab+bc+ca\right)\)\(\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=4\left[a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)\right]\)
\(\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=4\left(a^2b^2+b^2c^2+c^2a^2\right)\)
\(\Rightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2\right)=2\left(ab+bc+ca\right)^2\left(theoa\right)\)
Tìm x biết :
a) \(\left(x-2\right)^3+6\left(x+1\right)^2-x^3+12=0\)
b) \(\left(x-5\right)\left(x+5\right)-\left(x+3\right)^3+3\left(x-2\right)^2=\left(x+1\right)^2-\left(x+4\right)\left(x-4\right)+3x^2\)
c) \(\left(2x+3\right)^2+\left(x-1\right)\left(x+1\right)=5\left(x+2\right)^2-\left(x-5\right)\left(x+1\right)+\left(x+4\right)^2\)
d) \(\left(1-3x\right)^2-\left(x-2\right)\left(9x+1\right)=\left(3x-4\right)\left(3x+4\right)-9\left(x+3\right)^2\)
a/ \(x=\dfrac{-5}{12}\)
b/ \(x\approx-1,9526\)
c/ \(x=\dfrac{21-i\sqrt{199}}{10}\)
d/ \(x=\dfrac{-20}{13}\)
a) (x-2)3+6(x+1)2-x3+12=0
⇒ x3-6x2+12x-8+6(x2+2x+1)-x3+12=0
⇒ x3-6x2+12x-8+6x2+12x+6-x3+12=0
⇒ 24x+10=0
⇒ 24x=-10
⇒ x=-5/12
a.
PT \(\Leftrightarrow x^3-6x^2+12x-8+6(x^2+2x+1)-x^3+12=0\)
\(\Leftrightarrow x^3-6x^2+12x-8+6x^2+12x+6-x^3+12=0\)
\(\Leftrightarrow 24x+10=0\Leftrightarrow x=\frac{-5}{12}\)
b. Bạn xem lại đề, nghiệm khá xấu không phù hợp với mức độ tổng thể của bài.
c.
PT $\Leftrightarrow (4x^2+12x+9)+(x^2-1)=5(x^2+4x+4)+(x^2-4x-5)+9(x^2+6x+9)$
$\Leftrightarrow 10x^2+42x+64=0$
$\Leftrightarrow x^2+(3x+7)^2=-15< 0$ (vô lý)
Do đó pt vô nghiệm.
d.
PT $\Leftrightarrow (1-6x+9x^2)-(9x^2-17x-2)=(9x^2-16)-9(x^2+6x+9)$
$\Leftrightarrow 11x+3=-54x-97$
$\Leftrightarrow 65x=-100$
$\Leftrightarrow x=\frac{-20}{13}$