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Bài 1.
a) ( 7x - 3 )2 - 5x( 9x + 2 ) - 4x2 = 18
<=> 49x2 - 42x + 9 - 45x2 - 10x - 4x2 = 18
<=> -52x + 9 = 18
<=> -52x = 9
<=> x = -9/52
b) ( x - 7 )2 - 9( x + 4 )2 = 0
<=> x2 - 14x + 49 - 9( x2 + 8x + 16 ) = 0
<=> x2 - 14x + 49 - 9x2 - 72x - 144 = 0
<=> -8x2 - 86x - 95 = 0
<=> -8x2 - 10x - 76x - 95 = 0
<=> -8x( x + 5/4 ) - 76( x + 5/4 ) = 0
<=> ( x + 5/4 )( -8x - 76 ) = 0
<=> \(\orbr{\begin{cases}x+\frac{5}{4}=0\\-8x-76=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{4}\\x=-\frac{19}{2}\end{cases}}\)
c) ( 2x + 1 )2 + ( 4x - 1 )( x + 5 ) = 36
<=> 4x2 + 4x + 1 + 4x2 + 19x - 5 = 36
<=> 8x2 + 23x - 4 - 36 = 0
<=> 8x2 + 23x - 40 = 0
=> Vô nghiệm ( lớp 8 chưa học nghiệm vô tỉ nghen ) :))
Bài 2.
a) x2 - 12x + 39 = ( x2 - 12x + 36 ) + 3 = ( x - 6 )2 + 3 ≥ 3 > 0 ∀ x ( đpcm )
b) 17 - 8x + x2 = ( x2 - 8x + 16 ) + 1 = ( x - 4 )2 + 1 ≥ 1 > 0 ∀ x ( đpcm )
c) -x2 + 6x - 11 = -( x2 - 6x + 9 ) - 2 = -( x - 3 )2 - 2 ≤ -2 < 0 ∀ x ( đpcm )
d) -x2 + 18x - 83 = -( x2 - 18x + 81 ) - 2 = -( x - 9 )2 - 2 ≤ -2 < 0 ∀ x ( đpcm )
a) \(x^2-12x+11\)\(=0\)
\(\Leftrightarrow\left(x-6\right)^2-25=0\)
\(\Leftrightarrow\left(x-6+5\right)\left(x-6-5\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-11=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=11\end{matrix}\right.\)
a)\(x^2-12x+11=0\)
\(x^2-x-11x+11=0\)
\(\left(x^2-x\right)-\left(11x-11\right)=0\)
\(x\left(x-1\right)-11\left(x-1\right)=0\)
\(\left(x-1\right)\left(x-11\right)=0\)
\(=>\left[{}\begin{matrix}x-1=0\\x-11=0\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=1\\x=11\end{matrix}\right.\)
b)\(4x^2-4x-3=0\)
\(4x^2-2x+6x-3=0\)
\(2x\left(2x-1\right)+3\left(3x-1\right)=0\)
\(\left(2x-1\right)\left(2x+3\right)=0\)
\(=>\left[{}\begin{matrix}2x-1=0\\2x+3=0\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=0,5\\x=-1,5\end{matrix}\right.\)\
c)\(4x^2-12x-7=0\)
\(4x^2-14x+2x-7=0\)
\(2x\left(2x-7\right)+\left(2x-7\right)=0\)
\(\left(2x-7\right)\left(2x+1\right)=0\)
\(=>\left[{}\begin{matrix}2x-7=0\\2x+1=0\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=3,5\\x=-0,5\end{matrix}\right.\)
a) \(36x^2-49=0\)
\(\Leftrightarrow\left(6x\right)^2-7^2=0\)
\(\Leftrightarrow\left(6x-7\right)\left(6x+7\right)=0\)
\(TH_1:6x-7=0\) \(TH_2:6x+7=0\)
\(\Leftrightarrow6x=7\) \(\Leftrightarrow6x=-7\)
\(\Leftrightarrow x=\dfrac{7}{6}\) \(\Leftrightarrow x=-\dfrac{7}{6}\)
Vậy pt có tập nghiệm \(S=\left\{\dfrac{7}{6};-\dfrac{7}{6}\right\}\)
Bài 2
a) 36x2-49=0
⇔ (6x)2-49=0
⇔(6x-7).(6x+7)=0
TH1: 6x-7=0 TH2: 6x+7=0
⇔6x=7 ⇔6x=-7
⇔x=7/6 ⇔x=-7/6
Câu 1:
a) \((a+b)^3-3ab(a+b)=a^3+3a^2b+3ab^2+b^3-3ab(a+b)\)
\(=a^3+b^3+3ab(a+b)-3ab(a+b)\)
\(=a^3+b^3\)
Áp dụng: \(a^3+b^3=(a+b)^3-3ab(a+b)=(-5)^3-3.6(-5)=-35\)
b) \((a-b)^3+3ab(a-b)\)
\(=a^3-3a^2b+3ab^2-b^3+3ab(a-b)\)
\(=a^3-b^3-3ab(a-b)+3ab(a-b)\)
\(=a^3-b^3\)
Áp dụng:
\(a^3-b^3=(a-b)^3+3ab(a-b)=(-5)^3+3(-6)(-5)=-35\)
Câu 2:
a) Vì \(x^2\geq 0, \forall x\Rightarrow A=4x^2+3\geq 4.0+3=3\)
Vậy GTNN của $A$ là $3$ tại $x^2=0$ hay $x=0$
b)
\(B=2x^2+2x+2xy+y^2+3=(x^2+2x+1)+(x^2+2xy+y^2)+2\)
\(=(x+1)^2+(x+y)^2+2\)
Vì \((x+1)^2\geq 0; (x+y)^2\geq 0, \forall x,y\in\mathbb{R}\)
\(\Rightarrow B\geq 0+0+2=2\)
Vậy GTNN của $B$ là $2$ tại \(\left\{\begin{matrix} (x+1)^2=0\\ (x+y)^2=0\end{matrix}\right.\Leftrightarrow x=-1; y=1\)
Bài 1:
\(\left\{{}\begin{matrix}a=5c+1\\b=5d+2\end{matrix}\right.\)
\(a^2+b^2=\left(5c+1\right)^2+\left(5d+2\right)^2\)
\(=25c^2+10c+1+25d^2+20d+4\)
\(=25c^2+25d^2+10c+20d+5\)
\(=5\left(5c^2+5d^2+2c+4d+1\right)⋮5\)
Bài 3:
a: \(4x^2+12x+15=4x^2+12x+9+6=\left(2x+3\right)^2+6>=6\forall x\)
Dấu '=' xảy ra khi x=-3/2
b: \(9x^2-6x+5=9x^2-6x+1+4=\left(3x-1\right)^2+4>=4\forall x\)
Dấu '=' xảy ra khi x=1/3
a: =>2x^2-2x+2x-2-2x^2-x-4x-2=0
=>-5x-4=0
=>x=-4/5
b: =>6x^2-9x+2x-3-6x^2-12x=16
=>-19x=19
=>x=-1
c: =>48x^2-12x-20x+5+3x-48x^2-7+112x=81
=>83x=83
=>x=1
Vì a : 5 dư 2
b: 5 dư 3
\(\Rightarrow\) a; b lần lượt có dạng 5k+2; 5k+3
\(\Rightarrow\)ab=(5k+2).(5k+3)
=5k(5k+3)+2(5k+3)
=25k2+15k+10k+6
=25k2+25k+5+1
=5.(5k2+5k+1)+1
Ta có : \(5⋮5\)\(\Rightarrow5.\left(5k^2+5k+1\right)⋮5\)
Mà 1:5 =0 dư 1
\(\Rightarrow5.\left(5k^2+5k+1\right)+1:5 \left(d\text{ư}1\right)\)
\(\Rightarrow ab:5 \left(d\text{ư}1\right)\)
Điều phải chứng minh
Đặt a = 5k + 2. b = 5x + 3 ( do a chia 5 dư 2, b chia 5 dư 3 )
=> ab = (5k+2)(5x+3) = 25kx+10x+15k + 6
Ta có 25kx chia hết cho 5, 10x chia hết cho 5, 15k chia hết cho 5, 6 chia 5 dư 1 => ab chia 5 dư 1
Chúc bạn học tốt ^_^
a) 4x^2 -12x = (2x)^2 - 2.2x.3 + 3^2 - 3^2
= (2x-3)^2 - 3^2
= (2x - 3 -3)(2x-3 +3)
= 2x(2x - 6)
b) x^2 - y^2 -5x +5y = (x^2 - y^2) - (5x -5y)
= (x+y)(x-y) - 5(x-y)
= (x+y - 5)(x-y)
2. 3x(x - 5) -x +5 = 0
=>3x(x - 5) - (x -5) = 0
=> (3x - 1)(x-5) = 0
=>| 3x - 1 =0 => | 3x = 1 => |x = 1/3
| x - 5 =0 | x = 5 |x= 5
a) Ta có: \(Q=12x-4x^2-11=-\left(4x^2-12x+9\right)-2=-\left(2x-3\right)^2-2< 0\)
\(\Rightarrow\)ĐPCM
b) Ta có: \(\left(a^3+b^3\right)\left(a^2+b^2\right)=a^5+b^5+a^3b^2+a^2b^3=a^5+b^5+a^2b^2\left(a+b\right)\)
Mà \(ab=1\Rightarrow\left(a^3+b^3\right)\left(a^2+b^2\right)=a^5+b^5+\left(a+b\right)\)
\(\Rightarrow a^5+b^5=\left(a^3+b^3\right)\left(a^2+b^2\right)-\left(a+b\right)\)
a) We have \(Q=12x-4x^2-11=-\left(4x^2-12x+9\right)-2=-\left(2x-3\right)^2-2\)Because \(-\left(2x-3\right)^2\le0\); \(-\left(2x-3\right)^2-2\le-2< 0\Leftrightarrow Q< 0\)
And that's the thing we have to prove.
b) Just expand the polynomial on the right side of the equality:
We have \(R=\left(a^3+b^3\right)\left(a^2+b^2\right)-\left(a+b\right)\) \(=a^5+b^5+a^3b^2+a^2b^3-\left(a+b\right)\)\(=a^5+b^5+a^2b^2\left(a+b\right)-\left(a+b\right)\)
On the other hand, \(ab=1\Leftrightarrow a^2b^2=1\)
Therefore, \(R=a^5+b^5+\left(a+b\right)-\left(a+b\right)=a^5+b^5=L\)
Thus, the quality was proved.