Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1/
a. \(3x\left(5x^2-2x-1\right)\)
\(=15x^3-6x^2-3x\)
b. \(\left(x^2-2xy+3\right)\left(-xy\right)\)
\(=-x^3y+2x^2y^2-3xy\)
c. \(\left(2x^2-3xy+y^2\right)\left(x+y\right)\)
\(=2x^3-3x^2y+xy^2+2x^2y-3xy^2+y^3\)
\(=2x^3-x^2y-2xy^2\)
a) thiếu đề
b) \(\left(3x-3\right)\left(5-21x\right)+\left(7x+4\right)\left(9x-5\right)=44\)
\(15x-63x^2-15+63x+63x^2-35x+36x-20=44\)
\(79x-35=40\)
\(79x=75\)
\(x=\frac{75}{79}\)
\(\left(5\cdot\left(x^2-3x+1\right)+x\cdot\left(1-5x\right)\right)-\left(x-2\right)=0\)
\(7-15x=0\)
\(-15x=-7\)
\(x=\frac{7}{15}=0.467\)
\(b,\)câu b dài quá nên mik lười, vậy mik ghi kết quả thôi nhé
\(x=\frac{2}{19}=0.105\)
\(c,\)câu c cũng vậy mik ghi kết quả thôi nhé bn
\(x=-\frac{6}{11}=-0.545\)
1 ) Thực hiện phép tính :
a ) \(-\frac{1}{3}xz\left(-9xy+15yz\right)+3x^2\left(2yz^2-yz\right)\)
\(=3x^2yz-5xyz^2+6x^2yz^2-3x^2yz\)
\(=-5xyz^2+6x^2yz^2\)
b ) \(\left(x-2\right)\left(x^2-5x+1\right)-x\left(x^2+11\right)\)
\(=x^3-5x^2-x-2x^2+10x-2-x^3-11x\)
\(=-7x^2-2x-2-x^3\)
c ) \(\left(x^3+5x^2-2x+1\right)\left(x-7\right)\)
\(=x^4+5x^3-2x^2+x-7x^3-35x^2+14x-7\)
\(=x^4-2x^3-37x^2+15x-7\)
d ) \(\left(2x^2-3xy+y^2\right)\left(x+y\right)\)
\(=2x^3-3x^2y+xy^2+2x^2y-3xy^2+y^3\)
\(=2x^3-x^2y-2xy^2+y^3\)
e ) \(\left[\left(x^2-2xy+2y^2\right)\left(x+2y\right)-\left(x^2-4y^2\right)\left(x-y\right)\right]2xy\)
( để xem lại )
2 Tìm x
a ) \(6x\left(5x+3\right)+3x\left(1-10x\right)=7\)
\(\Leftrightarrow30x^2+18x+3x-30x^2=7\)
\(\Leftrightarrow21x=7\)
\(\Leftrightarrow x=3\)
b ) Sai đề
c ) \(\left(x+1\right)\left(x+2\right)\left(x+5\right)-x^2\left(x+8\right)=27\)
( Để xem lại )
mình chép đúng theo đề cô cho mà sao lại sai được ,hay cô cho sai đề
\(1,5\left(x^2-3x+1\right)+x\left(1-5x\right)=x-2\)
\(\Leftrightarrow5x^2-15x+5+x-5x^2-x+2=0\)
\(\Leftrightarrow-15x^2+7=0\)
\(\Leftrightarrow x^2=\frac{7}{15}\)
\(\Leftrightarrow x=\overset{+}{-}\sqrt{\frac{7}{15}}\)
\(2,7x\left(x-2\right)-5\left(x-1\right)=21x^2-14x^2+3\)
\(\Leftrightarrow7x^2-14x-5x+5-7x-3=0\)
\(\Leftrightarrow7x^2-26x+2=0\)
\(\Leftrightarrow7\left(x^2-\frac{26}{7}x+\frac{169}{49}\right)-\frac{155}{7}=0\)
\(\Leftrightarrow7\left(x-\frac{13}{7}\right)^2-\frac{155}{7}=0\)
\(\Leftrightarrow\left(\sqrt{7}x-\frac{13}{\sqrt{7}}-\sqrt{\frac{155}{7}}\right)\left(\sqrt{7}x-\frac{13}{\sqrt{7}}+\sqrt{\frac{155}{7}}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\sqrt{155}+13}{7}\\x=\frac{-\sqrt{155}+13}{7}\end{matrix}\right.\)
A = 3x2 - 5x + 1
= 3( x2 - 5/3x + 25/36 ) - 13/12
= 3( x - 5/6 )2 - 13/12 ≥ -13/12 ∀ x
Dấu "=" xảy ra khi x = 5/6
=> MinA = -13/12 <=> x = 5/6
B = 7x2 + 21x + 32
= 7( x2 + 3x + 9/4 ) + 65/4
= 7( x + 3/2 )2 + 65/4 ≥ 65/4 ∀ x
Dấu "=" xảy ra khi x = -3/2
=> MinB = 65/4 <=> x = -3/2
C = \(\frac{5}{x-x^2}\)
Để C đạt Min => x - x2 đạt Max
Ta có x - x2 = -( x2 - x + 1/4 ) + 1/4 = -( x - 1/2 )2 + 1/4 ≤ 1/4 ∀ x
Dấu "=" xảy ra khi x = 1/2
=> Max x - x2 = 1/4 khi x = 1/2
=> MinC = \(\frac{5}{\frac{1}{4}}=20\)khi x = 1/2
a) \(A=3x^2-5x+1=3\left(x^2-\frac{5}{3}x+\frac{25}{36}\right)-\frac{13}{12}\)
\(=3\left(x-\frac{5}{6}\right)^2-\frac{13}{12}\ge-\frac{13}{12}\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(A=-\frac{13}{12}\Leftrightarrow3\left(x-\frac{5}{6}\right)^2=0\)
\(\Rightarrow x=\frac{5}{6}\)
Vậy Min(A) = -13/12 khi x = 5/6
A = 3x2 - 5x + 1
=> A = 3x2 - 5x +\(\frac{25}{12}-\frac{13}{12}\)
=> A = 3 ( x -\(\frac{5}{6}\))2 -\(\frac{13}{12}\ge-\frac{13}{12}\)
Dấu "=" xảy ra <=> \(3\left(x-\frac{5}{6}\right)^2=0\Leftrightarrow x-\frac{5}{6}=0\Leftrightarrow x=\frac{5}{6}\)
Vậy minA = - 13/12 <=> x = 5/6
B = 7x2 + 21x + 32
=> B = 7x2 + 21x +\(\frac{63}{4}+\frac{65}{4}\)
=> B = 7 ( x +\(\frac{3}{2}\))2 +\(\frac{65}{4}\ge\frac{65}{4}\)
Dấu "=" xảy ra <=> \(7\left(x+\frac{3}{2}\right)^2=0\Leftrightarrow x+\frac{3}{2}=0\Leftrightarrow x=-\frac{3}{2}\)
Vậy minB = 65/4 <=> x = - 3/2
C = \(\frac{5}{x-x^2}=\frac{5}{-x^2+x-\frac{1}{4}+\frac{1}{4}}=\frac{5}{-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}}\)
Để C đạt GTNN thì - ( x -\(\frac{1}{2}\))2 +\(\frac{1}{4}\) đạt GTLN
Mà - ( x -\(\frac{1}{2}\))2 +\(\frac{1}{4}\le\frac{1}{4}\)
Dấu "=" xảy ra <=> \(-\left(x-\frac{1}{2}\right)^2=0\Leftrightarrow x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}\)
Vậy minC =\(\frac{5}{\frac{1}{4}}=20\)<=> x = 1/2
b) Ta có: \(B=7x^2+21x+32=7\left(x^2+3x+\frac{9}{4}\right)+\frac{65}{4}\)
\(=7\left(x+\frac{3}{2}\right)^2+\frac{65}{4}\ge\frac{65}{4}\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(B=\frac{65}{4}\Leftrightarrow7\left(x+\frac{3}{2}\right)^2=0\)
\(\Rightarrow x=-\frac{3}{2}\)
Vậy Min(B) = 65/4 khi x = -3/2
\(A=3x^2-5x+1=3\left(x^2-\frac{5}{3}x+\frac{1}{3}\right)\)
\(=3\left[x^2-2\cdot x\cdot\frac{5}{6}+\left(\frac{5}{6}\right)^2\right]-\frac{13}{12}\)
\(=3\left(x-\frac{5}{6}\right)^2-\frac{13}{12}\)
Vì \(\left(x-\frac{5}{6}\right)^2\ge0\forall x\)
=> \(3\left(x-\frac{5}{6}\right)^2-\frac{13}{12}\ge-\frac{13}{12}\forall x\)
Dấu " = " xảy ra khi và chỉ khi (x - 5/6)2 = 0 => x = 5/6
Vậy \(A_{min}=-\frac{13}{12}\)khi x = 5/6
\(B=7x^2+21x+32=7\left(x^2+3x+\frac{32}{7}\right)\)
\(=7\left[x^2+2\cdot x\cdot\frac{3}{2}+\left(\frac{3}{2}\right)^2\right]+\frac{65}{4}\)
\(=7\left(x+\frac{3}{2}\right)^2+\frac{65}{4}\)
Vì \(\left(x+\frac{3}{2}\right)^2\ge0\forall x\)
=> \(7\left(x+\frac{3}{2}\right)^2+\frac{65}{4}\ge\frac{65}{4}\forall x\)
Dấu " = " xảy ra khi và chỉ khi (x + 3/2)2 = 0 => x = -3/2
Vậy \(B_{min}=\frac{65}{4}\)khi x = -3/2
c) Đề bài là : 5x - x2 à ?
\(5x-x^2=-\left(x^2-5x\right)\)
\(=-\left[x^2-2\cdot x\cdot\frac{5}{2}+\left(\frac{5}{2}\right)^2\right]+\frac{25}{4}\)
\(=-\left(x-\frac{5}{2}\right)^2+\frac{25}{4}\)
Vì \(\left(x-\frac{5}{2}\right)^2\ge0\forall x\)
=> \(-\left(x-\frac{5}{2}\right)^2\le0\forall x\)
=> \(-\left(x-\frac{5}{2}\right)^2+\frac{25}{4}\le\frac{25}{4}\forall x\)
Dấu " = " xảy ra khi và chỉ khi -(x - 5/2)2 = 0 => x = 5/2
Vậy Cmax = 25/4 khi x = 5/2
Hay \(\frac{5}{x-x^2}=\frac{5}{-\left(x^2-x\right)}=\frac{5}{-\left[x^2-2\cdot x\cdot\frac{1}{2}+\left(\frac{1}{2}\right)^2\right]+\frac{1}{4}}=\frac{5}{-\left(x+\frac{1}{2}\right)^2+\frac{1}{4}}\)
Vì \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)
=> \(-\left(x+\frac{1}{2}\right)^2\le0\forall x\)
=> \(-\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\le\frac{1}{4}\forall x\)
=> \(\frac{5}{-\left(x+\frac{1}{2}\right)^2+\frac{1}{4}}\le\frac{5}{\frac{1}{4}}=20\)
Dấu " = " xảy ra khi và chỉ khi -(x + 1/2)2 = 0 => x = -1/2
Vậy Cmax = 20 khi x = -1/2
c) Ta có: \(C=\frac{5}{x-x^2}=\frac{5}{\frac{1}{4}-\left(x^2-x+\frac{1}{4}\right)}=\frac{5}{\frac{1}{4}-\left(x-\frac{1}{2}\right)^2}\)
Đến đây ta không thể xét được GTNN của C vì có các TH nếu \(\frac{1}{4}-\left(x-\frac{1}{2}\right)^2< 0\)
thì sẽ rất khó để tìm được GTNN
\(A=3x^2-5x+1\)
\(=3\left(x^2-2x\frac{5}{6}+\frac{25}{36}\right)-\frac{13}{12}\)
\(=3\left(x-\frac{5}{6}\right)^2-\frac{13}{12}\ge-\frac{13}{12}\forall x\)
Dấu"="xảy ra khi \(\left(x-\frac{5}{6}\right)^2=0\Rightarrow x=\frac{5}{6}\)
Vậy \(Min_A=-\frac{13}{12}\Leftrightarrow x=\frac{5}{6}\)
b,\(B=7x^2+21x+32\)
\(=7\left(x^2+3x+\frac{32}{7}\right)\)
\(=7\left(x^2+2x\frac{3}{2}+\frac{9}{4}\right)+\frac{65}{4}\)
\(=7\left(x+\frac{3}{2}\right)^2+\frac{65}{4}\ge\frac{65}{4}\forall x\)
Tự lm tiếp