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Bài 1 quan trong là đoán dấu đẳng thức.
1/ Có: \(36=\left(3+2+1\right)\left(a^2+b^2+c^2\right)\ge\left(\sqrt{3}a+\sqrt{2}b+c\right)^2\)
\(\therefore\sqrt{3}a+\sqrt{2}b+c\le6\)
\(\frac{1}{3}\left(\frac{a}{bc}+\frac{3b}{2ca}\right)+\frac{3}{2}\left(\frac{b}{ca}+\frac{2c}{ab}\right)+2\left(\frac{c}{ab}+\frac{a}{3bc}\right)\)
\(\ge\frac{\sqrt{6}}{3c}+\frac{3\sqrt{2}}{a}+\frac{4\sqrt{3}}{3b}\)
\(=\frac{\left(\frac{\sqrt{6}}{3}\right)}{c}+\frac{\left(3\sqrt{6}\right)}{\sqrt{3}a}+\frac{\left(\frac{4\sqrt{6}}{3}\right)}{\sqrt{2}b}\)
\(\ge\frac{\left(\sqrt{\frac{\sqrt{6}}{3}}+\sqrt{3\sqrt{6}}+\sqrt{\frac{4\sqrt{6}}{3}}\right)^2}{\sqrt{3}a+\sqrt{2}b+c}\ge2\sqrt{6}\)
Đẳng thức xảy ra khi \(a=\sqrt{3},b=\sqrt{2},c=1\)
a,\(A=\left(x^4-3x^2+9\right)\left(x^2+3\right)+\left(3-x^2\right)^2\)
\(A=x^6-3x^4+9x^2+3x^4-9x^2+27+9-6x^2+x^4\)
\(A=x^6+x^4-6x^2+36\)
b, \(M=5\left(x+2y\right)^2-\left(3y+2x\right)^2+\left(4x-y\right)^2+3\left(x-2y\right)\left(x+2y\right)\)
\(M=5\left(x^2+4xy+4y^2\right)-\left(9y^2+12xy+4x^2\right)+\left(16x^2-8xy+y^2\right)+3\left(x^2-4y^2\right)\)
\(M=5x^2+20xy+20y^2-9y^2-12xy-4x^2+16x^2-8xy+y^2+3x^2-12y^2\)
\(M=20x^2\)
Các câu còn lại làm tương tự! Chúc bạn học tốt!!!
E=\(\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(1+6x\right)\left(6x-1\right)\)
\(\Leftrightarrow\left(6x+1\right)^2-2\left(1+6x\right)\left(6x-1\right)+\left(6x-1\right)^2\)
\(\Leftrightarrow\left[\left(6x+1\right)-\left(6x-1\right)\right]^2\)
\(\Leftrightarrow\left(6x+1-6x+1\right)^2=2^2=4\)
Bài 2:
a: \(3\left(x-1\right)\left(x^2+x+1\right)+\left(x-1\right)^3-4x\left(x+1\right)\left(x-1\right)\)
\(=3\left(x^3-1\right)+x^3-3x^2+3x-1-4x\left(x^2-1\right)\)
\(=3x^3-3+x^3-3x^2+3x-1-4x^3+4x\)
\(=-3x^2+7x-4\)
\(=-3\cdot\left(-1\right)^2+7\cdot\left(-1\right)-4\)
=-3-4-7=-14
b: \(=27x^3y^3-8-3xy\left(9x^2y^2+6xy+1\right)\)
\(=27x^3y^3-8-27x^3y^3-18x^2y^2-3xy\)
\(=-18x^2y^2-3xy-8\)
\(=-18\cdot\left[\left(-2010\right)\cdot\left(-\dfrac{1}{2010}\right)\right]^2-3\cdot\left(-2010\right)\cdot\dfrac{-1}{2010}-8\)
\(=-18-3-8=-29\)
Bài 1:
a) \(8\left(x-2\right)-2\left(3x-4\right)=2\)
\(\Rightarrow2\left[4\left(x-2\right)-\left(3x-4\right)\right]=2\)
\(\Rightarrow4\left(x-2\right)-3x+4=0\)
\(\Rightarrow4x-8-3x+4=0\)
\(\Rightarrow x-4=0\)
\(\Rightarrow x=4\)
b) \(10\left(3x-2\right)-3\left(5x+2\right)+5\left(11-4x\right)=25\)
\(\Rightarrow5\left[2\left(3x-2\right)+11-4x\right]-3\left(5x+2\right)=25\)
\(\Rightarrow5\left(6x-4+11-4x\right)-3\left(5x+2\right)=25\)
\(\Rightarrow5\left(2x+7\right)-3\left(5x+2\right)=25\)
\(\Rightarrow10x+35-15x-6=25\)
\(\Rightarrow-5x+29=25\)
\(\Rightarrow-5x=25-29\)
\(\Rightarrow-5x=-4\)
\(\Rightarrow x=\dfrac{4}{5}\)
c) \(2x\left(x+1\right)-x^2\left(x+2\right)+x^3-x+4=0\)
\(\Rightarrow2x^2+2x-x^3-2x^2+x^3-x+4=0\)
\(\Rightarrow x+4=0\)
\(\Rightarrow x=-4\)
d) \(4x\left(3x+2\right)-6x\left(2x+5\right)+21\left(x-1\right)=0\)
\(\Rightarrow12x^2+8x-12x^2-30x+21x-21=0\)
\(\Rightarrow-x-21=0\)
\(\Rightarrow x=-21\)
Bài 2:
a) \(P=\left(4x^2-3y\right)2y-\left(3x^2-4y\right)3y\)
\(P=8x^2y-6y^2-9x^2y+12y^2\)
\(P=-x^2y+6y^2\)
Thay x = -1 ; y = 2 vào P ta được
\(P=-\left(-1\right)^2.2+6.2^2\)
\(P=-2+24=22\)
b) \(Q=4x^2\left(5x-3y\right)-x^2\left(4x+y\right)\)
\(Q=20x^3-12x^2y-4x^3-x^2y\)
\(Q=16x^3-13x^2y\)
Thay x = -1 ; y = 2 vào Q ta được
\(Q=16\left(-1\right)^3-13\left(-1\right)^2.2\)
\(Q=-16-26\)
\(Q=-42\)
c) \(H=x\left(x^3-y\right)+x^2\left(y-x^2\right)-y\left(x^2-3x\right)\)
\(H=x^4-xy+x^2y-x^4-x^2y+3xy\)
\(H=2xy\)
Thay x = 1/4 ; y = 2012 vào H ta được
\(H=2.\dfrac{1}{4}.2012\)
\(H=1006\)
1.a)\(8\left(x-2\right)-2\left(3x-4\right)=2\)
\(\Leftrightarrow8x-16-6x+8=2\)
\(\Leftrightarrow2x-8=2\Leftrightarrow2x=10\Leftrightarrow x=5\)
b)\(10\left(3x-2\right)-3\left(5x+2\right)+5\left(11-4x\right)=25\)
\(\Leftrightarrow30x-20-15x-6+55-20x=25\)
\(\Leftrightarrow-5x+29=25\Leftrightarrow-5x=-4\Leftrightarrow x=\dfrac{4}{5}=0,8\)
\(c)2x\left(x+1\right)-x^2\left(x+2\right)+x^3-x+4=0\)
\(\Leftrightarrow2x^2+2x-x^3-2x^2+x^3-x+4=0\)
\(\Leftrightarrow x+4=0\Leftrightarrow x=-4\)
\(d)4x\left(3x+2\right)-6x\left(2x+5\right)+21\left(x-1\right)=0\)
\(\Leftrightarrow12x^2+8x-12x^2-30x+21x-21=0\)
\(\Leftrightarrow-x-21=0\Leftrightarrow-x=21\Leftrightarrow x=-21\)
2.
a)\(P=\left(4x^2-3y\right)2y-\left(3x^2-4y\right)3y\)
\(\Leftrightarrow8x^2y-6y^2-9x^2y-12y^2\)
\(\Leftrightarrow x^2y-18y^2\)
tại x=-1 , y=2
ta có:\(x^2y-18y^2=\left(-1\right)^2.2-18.2^2=2-72=-70\)
vậy \(P=\left(4x^2-3y\right)2y-\left(3x^2-4y\right)3y=-70\) tại x=-1,y=2
b)\(Q=4x^2\left(5x-3y\right)-x^2\left(4x+y\right)\)
\(\Leftrightarrow20x^3-12x^2y-4x^3-x^2y\)
\(\Leftrightarrow17x^3-13x^2y\)
tại x=-1,y=2
ta có:\(17x^3-13x^2y=17\left(-1\right)^3-13\left(-1\right)^2.2=-17-26=-43\)
vậy \(Q=4x^2\left(5x-3y\right)-x^2\left(4x+y\right)=-43\)
c)\(H=x\left(x^3-y\right)+x^2\left(y-x^2\right)-y\left(x^2-3x\right)\)
\(\Leftrightarrow x^4-xy+x^2y-x^3-x^2y+3xy\)
\(\Leftrightarrow x^4+2xy-x^3\)
tại x=1/4 và y=2012
ta có:\(x^4+2xy-x^3=\left(\dfrac{1}{4}\right)^4+2.\dfrac{1}{4}.2012-\left(\dfrac{1}{4}\right)^3\approx1006\)
Bài 1.
a)
\((x-2)(2x-1)-(2x-3)(x-1)-2\\=2x^2-x-4x+2-(2x^2-2x-3x+3)-2\\=2x^2-5x+2-(2x^2-5x+3)-2\\=2x^2-5x+2-2x^2+5x-3-2\\=(2x^2-2x^2)+(-5x+5x)+(2-3-2)\\=-3\)
b)
\(x(x+3y+1)-2y(x-1)-(y+x+1)x\\=x^2+3xy+x-2xy+2y-xy-x^2-x\\=(x^2-x^2)+(3xy-2xy-xy)+(x-x)+2y\\=2y\)
Bài 2.
a)
\((14x^3+12x^2-14x):2x=(x+2)(3x-4)\\\Leftrightarrow 14x^3:2x+12x^2:2x-14x:2x=3x^2-4x+6x-8\\ \Leftrightarrow 7x^2+6x-7=3x^2+2x-8\\\Leftrightarrow (7x^2-3x^2)+(6x-2x)+(-7+8)=0\\\Leftrightarrow 4x^2+4x+1=0\\\Leftrightarrow (2x)^2+2\cdot 2x\cdot 1+1^2=0\\\Leftrightarrow (2x+1)^2=0\\\Leftrightarrow 2x+1=0\\\Leftrightarrow 2x=-1\\\Leftrightarrow x=\frac{-1}2\)
b)
\((4x-5)(6x+1)-(8x+3)(3x-4)=15\\\Leftrightarrow 24x^2+4x-30x-5-(24x^2-32x+9x-12)=15\\\Leftrightarrow 24x^2-26x-5-(24x^2-23x-12)=15\\\Leftrightarrow 24x^2-26x-5-24x^2+23x+12=15\\\Leftrightarrow -3x+7=15\\\Leftrightarrow -3x=8\\\Leftrightarrow x=\frac{-8}3\\Toru\)
Bài 2;
\(a)x^4-16x=0\Rightarrow x^4=16x\Leftrightarrow x^3=16\Leftrightarrow x=\sqrt[3]{16}\)
\(c)4x^2-\frac{1}{4}=0\Leftrightarrow4x^2=\frac{1}{4}\Leftrightarrow x^2=\frac{1}{16}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{4}\\x=-\frac{1}{4}\end{cases}}\)
1)\(21x^2y-12xy^2=xy.\left(21x-12y\right)\)
2)\(x^3+x^2-2x=x.\left(x^2+x-2\right)\)
3)\(3x.\left(x-1\right)+7x^2\left(x-1\right)=\left(x-1\right).\left(3x+7x^2\right)=x.\left(x-1\right)\left(3+7x\right)\)
15)\(\left(2a+3\right)^2-\left(2a+1\right)^2=\left(2a+3-2a-1\right)\left(2a+3+2a+1\right)=2.\left(4a+4\right)=8\left(a+1\right)\)
14) \(-4y^2+4y-1=-\left[\left(2y\right)^2-2.2y.1+1^2\right]=-\left(2y-1\right)^2\)
13) \(x^6+1=\left(x^2\right)^3+1=\left(x^2+1\right)\left(x^4-x^2+1\right)\)
12) \(\left(x+1\right)^2-\left(y+6\right)^2=\left(x+1-y-6\right)\left(x+1+y+6\right)=\left(x-y-5\right)\left(x+y+7\right)\)
4) \(3x\left(x-a\right)+4a\left(a-x\right)=3x.\left(x-a\right)-4a\left(x-a\right)=\left(x-a\right)\left(3x-4a\right)\)
Sao nhiều thế!
câu 1 là : Tìm x để A khác 0 \(A=\frac{-4x^2}{3-x}\)