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Bài 2 :
Câu a : \(y\left(y^3+y^2-y-2\right)-\left(y^2-2\right)\left(y^2+y+1\right)\)
\(=y^4+y^3-y^2-2y-y^4-y^3-y^2+2y^2+2y+2\)
\(=2\) \(\Rightarrow\) ko phụ thuộc vào biến .
Câu b : \(\left(2x+3\right)\left(4x^2-6x+9\right)-2\left(4x^3-1\right)\)
\(=8x^3-12x^2+18x+12x^2-18x+27-8x^3+2\)
\(=29\Rightarrow\) ko thuộc vào biến
Câu c : \(3x\left(x+5\right)-\left(3x+18\right)\left(x-1\right)\)
\(=3x^2+15x-3x^2+3x-18x+18\)
\(=18\) \(\Rightarrow\) ko thuộc vào biến
Câu d : \(\left(2x+6\right)\left(4x^2-12x+36\right)-8x^3+5\)
\(=8x^3-24x^2+72x+24x^2-72x+216-8x^3+5\)
\(=221\) \(\Rightarrow\) không thuộc vào biến
câu 1) a) \(\left(x^2+2xy+y^2\right)\left(x+y\right)=\left(x+y\right)^2\left(x+y\right)=\left(x+y\right)^3\)
b) \(y\left(y^3+y^2-3y-2\right)+\left(y^2-2\right)\left(y^2+y-1\right)\)
\(=y^4+y^3-3y^2-2y+y^4+y^3-y^2-2y^2-2y+2\)
\(=2y^4+2y^3-6y^2-4y+2=2y\left(y^3+y^2-3y-2\right)+2\)
\(=2y\left(y+2\right)\left(y^2-y-1\right)+2=2\left(y^2+2y\right)\left(y^2-y-1\right)+2\)
\(=2\left(y^2+2y\right)\left(y^2-y-1+1\right)=2\left(y^2+2y\right)\left(y^2-y\right)\)
c) \(6x^2-\left(2x+5\right)\left(3x-2\right)=6x^2-\left(6x^2-4x+15x-10\right)\)
\(\Leftrightarrow6x^2-6x^2+4x-15x+10=-11x+10\)
d) \(\left(2x-1\right)\left(3x+1\right)+\left(3x+4\right)\left(3-2x\right)\)
\(\)\(=6x^2+2x-3x-1+9x-6x^2+12-8x=11\)
e) \(\left(3x-5\right)\left(7-5x\right)-\left(5x+2\right)\left(2-3x\right)\)
\(=21x-15x^2-35+25x-\left(10x-15x^2+4-6x\right)\)
\(21x-15x^2-35+25x-10x+15x^2-4+6x=42x-39\)
Bài1:
\(4\left(x+1\right)^2+\left(2x-1\right)^2+8\left(x+1\right)\left(x-1\right)=11\)
=>\(4\left(x^2+2x+1\right)+\left(4x^2-4x+1\right)+8\left(x^2-1\right)=11\)
=>\(4x^2+8x+4+4x^2-4x+1+8x^2-8=11\)
=>\(4x=14\)
=>\(x=\dfrac{7}{2}\)
Vậy..
Các câu sau tương tự
Bài2:
\(a,A=x^2-2x+10\)
=\(\left(x-1\right)^2+9\)
Với ọi x thì \(\left(x-1\right)^2+9\ge9\)
Hay \(A\ge9\)
Để A=9 thì\(\left(x-1\right)^2=0\Leftrightarrow x-1=0\Leftrightarrow x=1\)
Vậy
Các câu sau tương tự
1.
a. \((x+1)(x^2-x+1)-(x-1)(x^2+x+1)\)
\(=x^3 + 1-(x^3-1) = 2 \)
b.
\(\dfrac{2x^2-4x+2}{2x-2}=\dfrac{2\left(x^2-2x+1\right)}{2\left(x-1\right)}=\dfrac{\left(x-1\right)^2}{x-1}=x-1\)
2.
a. \(x^2-4y^2+12y-9=x^2-\left[\left(2y\right)^2-2\cdot2y\cdot3+3^2\right]=x^2-\left(2y-3\right)^2=\left(x-2y+3\right)\left(x+2y-3\right)\)
b.
\(5x^2+3\left(x+y\right)^2-5y^2\)
\(=3\left(x+y\right)^2+5\left(x^2-y^2\right)\)
\(=3\left(x+y\right)^2+5\left(x+y\right)\left(x-y\right)\)
\(=\left(x+y\right)\left[3\left(x+y\right)+5\left(x-y\right)\right]\)
\(=\left(x+y\right)\left(3x+3y+5x-5y\right)\)
\(=\left(x+y\right)\left(8x-2y\right)=2\left(x+y\right)\left(4x-y\right)\)
\(a.17+8x=10-6x\\\Leftrightarrow 8x+6x=-17+10\\\Leftrightarrow 2x=-7\\ \Leftrightarrow x=-\frac{7}{2}\)
Vậy nghiệm của phương trình trên là \(-\frac{7}{2}\)
\(b.3\left(x+5\right)+7=19-5\left(x-2\right)\\\Leftrightarrow 3x+15+7=19-5x+10\\ \Leftrightarrow3x+5x=-15-7+19+10\\ \Leftrightarrow8x=7\\\Leftrightarrow x=\frac{7}{8}\)
Vậy nghiệm của phương trình trên là \(\frac{7}{8}\)
\(c.3x-4\left(x+2\right)\left(x+3\right)=14-4\left(x^2-3x\right)\\ \Leftrightarrow3x-4\left(x^2+5x+6\right)=14-4x^2+12x\\ \Leftrightarrow4x^2-4x^2+3x-5x-12x=24+14\\ \Leftrightarrow-14x=38\\ \Leftrightarrow x=-\frac{19}{7}\)
Vậy nghiệm của phương trình trên là \(-\frac{19}{7}\)
\(d.x+\frac{3}{4}+3x+2=\frac{x}{3}-3x-\frac{2}{6}\\ \Leftrightarrow\frac{12x}{12}+\frac{9}{12}+\frac{36x}{12}+\frac{24}{12}=\frac{4x}{12}-\frac{36x}{12}-\frac{4}{12}\\ \Leftrightarrow12x+9+36x+24=4x-36x-4\\ \Leftrightarrow12x+36x+36x-4x=-24-9-4\\ \Leftrightarrow80x=-37\\ \Leftrightarrow x=-\frac{37}{80}\)
a) (x + 2)(x + 3) - (x - 2)(x + 5) = 0
<=> x2 + 3x + 2x + 6 - (x2 + 5x - 2x - 10) = 0
<=> x2 + 3x + 2x + 6 - x2 - 5x + 2x + 10 = 0
<=> 2x + 16 = 0
<=> 2x = -16
<=> x = -8
b) (2x + 3)(x - 4) + (x - 5)(x - 2) = (3x - 5)(x - 4)
<=> (2x + 3)(x - 4) + (x - 5)(x - 2) - (3x - 5)(x - 4) = 0
<=> 2x2 - 8x + 3x - 12 + x2 - 2x - 5x + 10 - (3x2 - 12x - 5x + 20) = 0
<=> 2x2 - 8x + 3x - 12 + x2 - 2x - 5x + 10 - 3x2 + 12x + 5x - 20 = 0
<=> 5x = 12 - 10 + 20
<=> 5x = 22
<=> x = 22/5
c) (8 - 5x)(x + 2) + 4(x - 2)(x + 1) + 2(x - 2)(x + 2) = 0
<=> 8x + 16 - 5x2 - 10x + (4x - 8)(x + 1) + 2(x2 - 4) = 0
<=> 8x + 16 - 5x2 - 10x + 4x2 + 4x - 8x - 8 + 2x2 - 8 = 0
<=> x2 - 6x = 0
<=> x(x - 6) = 0
<=> x = 0 hay x - 6 = 0
I<=> x = 6
d) (8x - 3)(3x + 2) - (4x + 7)(x + 4) = (2x + 1)(5x - 1) - 33
<=> 24x2 + 16x - 9x - 6 - (4x2 + 16x + 7x + 28) = 10x2 - 2x + 5x - 1 - 33
<=> 24x2 + 16x - 9x - 6 - 4x2 - 16x - 7x - 28 - 10x2 + 2x - 5x + 1 + 33 = 0
<=> 10x2 - 19x = 0
<=> x(10x - 19) = 0
<=> x = 0 hay 10x - 19 = 0
I <=> 10x = 19
I <=> x = 19/10
Bài 2:
b: \(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-7x+10=3x^2-12x-5x+20\)
\(\Leftrightarrow3x^2-12x-2=3x^2-17x+20\)
=>-12x-2=-17x+20
=>5x=22
hay x=22/5
c: \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)
\(\Leftrightarrow24x^2+16x-9x-6-\left(4x^2+16x+7x+28\right)=10x^2-2x+5x-1\)
\(\Leftrightarrow24x^2+7x-6-4x^2-23x-28=10x^2+3x-1\)
\(\Leftrightarrow20x^2-16x-34=10x^2+3x-1\)
\(\Leftrightarrow10x^2-19x-33=0\)
\(\text{Δ}=\left(-19\right)^2-4\cdot10\cdot\left(-33\right)=1681>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{19-41}{20}=\dfrac{-22}{20}=\dfrac{-11}{10}\\x_2=\dfrac{19+41}{20}=3\end{matrix}\right.\)
a ) \(VT=\left|x-1\right|+\left|2-x\right|\ge\left|x-1+2-x\right|=1=VP\)
Đẳng thức xảy ra khi \(\left(x-1\right)\left(2-x\right)\ge0\Rightarrow1\le x\le2\)
c ) \(VT=\left|x+1\right|+\left|2x+4\right|\ge\left|x+1+2x+4\right|=\left|3x+5\right|\ge3x+5=VP\)
Đẳng thức xảy ra khi \(\begin{cases}\left(x+1\right)\left(2x+4\right)\ge0\\3x+5\ge0\end{cases}\Rightarrow x\ge1\)