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Bài 1.
a) x( 8x - 2 ) - 8x2 + 12 = 0
<=> 8x2 - 2x - 8x2 + 12 = 0
<=> 12 - 2x = 0
<=> 2x = 12
<=> x = 6
b) x( 4x - 5 ) - ( 2x + 1 )2 = 0
<=> 4x2 - 5x - ( 4x2 + 4x + 1 ) = 0
<=> 4x2 - 5x - 4x2 - 4x - 1 = 0
<=> -9x - 1 = 0
<=> -9x = 1
<=> x = -1/9
c) ( 5 - 2x )( 2x + 7 ) = ( 2x - 5 )( 2x + 5 )
<=> -4x2 - 4x + 35 = 4x2 - 25
<=> -4x2 - 4x + 35 - 4x2 + 25 = 0
<=> -8x2 - 4x + 60 = 0
<=> -8x2 + 20x - 24x + 60 = 0
<=> -4x( 2x - 5 ) - 12( 2x - 5 ) = 0
<=> ( 2x - 5 )( -4x - 12 ) = 0
<=> \(\orbr{\begin{cases}2x-5=0\\-4x-12=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-3\end{cases}}\)
d) 64x2 - 49 = 0
<=> ( 8x )2 - 72 = 0
<=> ( 8x - 7 )( 8x + 7 ) = 0
<=> \(\orbr{\begin{cases}8x-7=0\\8x+7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{8}\\x=-\frac{7}{8}\end{cases}}\)
e) ( x2 + 6x + 9 )( x2 + 8x + 7 ) = 0
<=> ( x + 3 )2( x2 + x + 7x + 7 ) = 0
<=> ( x + 3 )2 [ x( x + 1 ) + 7( x + 1 ) ] = 0
<=> ( x + 3 )2( x + 1 )( x + 7 ) = 0
<=> x = -3 hoặc x = -1 hoặc x = -7
g) ( x2 + 1 )( x2 - 8x + 7 ) = 0
Vì x2 + 1 ≥ 1 > 0 với mọi x
=> x2 - 8x + 7 = 0
=> x2 - x - 7x + 7 = 0
=> x( x - 1 ) - 7( x - 1 ) = 0
=> ( x - 1 )( x - 7 ) = 0
=> \(\orbr{\begin{cases}x-1=0\\x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=7\end{cases}}\)
Bài 2.
a) ( x - 1 )2 - ( x - 2 )( x + 2 )
= x2 - 2x + 1 - ( x2 - 4 )
= x2 - 2x + 1 - x2 + 4
= -2x + 5
b) ( 3x + 5 )2 + ( 26x + 10 )( 2 - 3x ) + ( 2 - 3x )2
= 9x2 + 30x + 25 - 78x2 + 22x + 20 + 9x2 - 12x + 4
= ( 9x2 - 78x2 + 9x2 ) + ( 30x + 22x - 12x ) + ( 25 + 20 + 4 )
= -60x2 + 40x2 + 49
d) ( x + y )2 - ( x + y - 2 )2
= [ x + y - ( x + y - 2 ) ][ x + y + ( x + y - 2 ) ]
= ( x + y - x - y + 2 )( x + y + x + y - 2 )
= 2( 2x + 2y - 2 )
= 4x + 4y - 4
Bài 3.
A = 3x2 + 18x + 33
= 3( x2 + 6x + 9 ) + 6
= 3( x + 3 )2 + 6 ≥ 6 ∀ x
Đẳng thức xảy ra <=> x + 3 = 0 => x = -3
=> MinA = 6 <=> x = -3
B = x2 - 6x + 10 + y2
= ( x2 - 6x + 9 ) + y2 + 1
= ( x - 3 )2 + y2 + 1 ≥ 1 ∀ x,y
Đẳng thức xảy ra <=> \(\hept{\begin{cases}x-3=0\\y^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=0\end{cases}}\)
=> MinB = 1 <=> x = 3 ; y = 0
C = ( 2x - 1 )2 + ( x + 2 )2
= 4x2 - 4x + 1 + x2 + 4x + 4
= 5x2 + 5 ≥ 5 ∀ x
Đẳng thức xảy ra <=> 5x2 = 0 => x = 0
=> MinC = 5 <=> x = 0
D = -2/7x2 - 8x + 7 ( sửa thành tìm Max )
Để D đạt GTLN => 7x2 - 8x + 7 đạt GTNN
7x2 - 8x + 7
= 7( x2 - 8/7x + 16/49 ) + 33/7
= 7( x - 4/7 )2 + 33/7 ≥ 33/7 ∀ x
Đẳng thức xảy ra <=> x - 4/7 = 0 => x = 4/7
=> MaxC = \(\frac{-2}{\frac{33}{7}}=-\frac{14}{33}\)<=> x = 4/7
Bài 1:
a) (3x-2).(4x+5)-6x.(2x-1) = 12x^2 +15x - 8x -10 - 12x^2 + 6x = 13x - 10
b) (2x-5)^2 - 4.(x+3).(x-3) = 4x^2 - 20x + 25 - 4x^2 + 12x -12x + 36 = -20x + 61
Bài 2:
a)(2x-1)^2-(x+3)^2 = 0
<=> (2x-1-x-3).(2x-1+x+3) =0
<=>(x-4).(3x+2) = 0
<=> x-4 = 0 hoặc 3x+2=0
*x-4=0 => x=4
*3x+2 = 0 => 3x=-2 => x=-2/3
b)x^2(x-3)+12-4x=0 <=> x^2(x-3) - 4(x-3) =0 <=> (x-3).(x-2)(x+2) <=> x-3=0 hoặc x-2=0 hoặc x+2 =0
*x-3=0 => x=3
*x-2=0 =>x=2
*x+2=0 =>x=-2
c) 6x^3 -24x =0 <=> 6x(x^2 -4)=0 <=> 6x(x-2)(x+2)=0 <=> x=0 hoặc x-2 =0 hoặc x+2=0 <=> x=0 hoặc x=2 hoặc x=-2
1) \(2\left(x+2\right)-\left(3x+1\right)\left(x+2\right)=0\)
\(\left(x+2\right)\left(2-3x-1\right)=0\)
\(\left(x+2\right)\left(1-3x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+2=0\\1-3x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-2\\x=\frac{1}{3}\end{cases}}}\)
2) \(3x\left(x-3\right)-\left(2x-6\right)=0\)
\(3x\left(x-3\right)-2\left(x-3\right)=0\)
\(\left(x-3\right)\left(3x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\3x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{2}{3}\end{cases}}}\)
3) \(\left(2x-1\right)^2=\left(3x-5\right)^2\)
\(\left(2x-1\right)^2-\left(3x-5\right)^2=0\)
\(\left(2x-1-3x+5\right)\left(2x-1+3x-5\right)=0\)
\(\left(4-x\right)\left(5x-6\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4-x=0\\5x-6=0\end{cases}\Rightarrow\orbr{\begin{cases}x=4\\x=\frac{6}{5}\end{cases}}}\)
4) \(\left(4x+3\right)\left(x-1\right)=x^2-1\)
\(\left(4x+3\right)\left(x-1\right)=\left(x+1\right)\left(x-1\right)\)
\(\left(4x+3\right)\left(x-1\right)-\left(x+1\right)\left(x-1\right)=0\)
\(\left(x-1\right)\left(4x+3-x-1\right)=0\)
\(\left(x-1\right)\left(3x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\3x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{-2}{3}\end{cases}}}\)
5) \(6-4x-\left(2x-3\right)\left(x-3\right)=0\)
\(-2\left(2x-3\right)-\left(2x-3\right)\left(x-3\right)=0\)
\(\left(2x-3\right)\left(-2-x+3\right)=0\)
\(\left(2x-3\right)\left(1-x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-3=0\\1-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=1\end{cases}}}\)
6) \(2x^2-5x-7=0\)
\(2x^2+2x-7x-7=0\)
\(2x\left(x+1\right)-7\left(x+1\right)=0\)
\(\left(x+1\right)\left(2x-7\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+1=0\\2x-7=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=\frac{7}{2}\end{cases}}}\)
7) \(x^2-x-12=0\)
\(x^2+3x-4x-12=0\)
\(x\left(x+3\right)-4\left(x+3\right)\)
\(\left(x+3\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\x-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-3\\x=4\end{cases}}}\)
8) \(3x^2+14x-5=0\)
\(3x^2+15x-x-5=0\)
\(3x\left(x+5\right)-\left(x+5\right)=0\)
\(\left(x+5\right)\left(3x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+5=0\\3x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-5\\x=\frac{1}{3}\end{cases}}}\)
6) c) x3 - x2 + x = 1
<=> x3 - x2 + x - 1 = 0
<=> (x3 - x2) + (x - 1) = 0
<=> x2 (x - 1) + (x - 1) = 0
<=> (x - 1) (x2 + 1) = 0
=> x - 1 = 0 hoặc x2 + 1 = 0
* x - 1 = 0 => x = 1
* x2 + 1 = 0 => x2 = -1 => x = -1
Vậy x = 1 hoặc x = -1
Bài 5:
a) Đặt \(A=\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=3^{32}-1\)
\(\Rightarrow A=\frac{3^{32}-1}{8}\)
b) (7x+6)2 + (5-6x)2 - (10-12x)(7x+6)
=(7x+6)2 + (5-6x)2 - 2(5-6x)(7x+6)
\(=\left(7x+6-5+6x\right)^2\)
\(=\left(13x+1\right)^2\)
Giải hết không nổi =.= đành giải vài bài thôi :v . Lần sau bạn nên đăng từ từ để người giải bớt ngán nhé!
Bài 1
a) \(2\left(x+5\right)=x^2+5x\)
\(\Leftrightarrow2x+10=x^2+5x\)
\(\Leftrightarrow x^2+5x-2x=10\)
\(\Leftrightarrow x^2+3x=10\Leftrightarrow x\left(x+3\right)=10\Leftrightarrow\hept{\begin{cases}x=-5\\x=2\end{cases}}\) (ở đây lười kẻ bảng quá =((( )
b) \(x\left(x-2\right)+x-2=0\)
\(\Leftrightarrow x^2-2x+x=2\Leftrightarrow x^2-x=2\)
\(\Leftrightarrow x\left(x-1\right)=2\Leftrightarrow\hept{\begin{cases}x=-1\\x=2\end{cases}}\) (bạn kẻ bảng ra các ước của 2 là thấy)
:v lời giải bài 1 đang chờ duyệt. Mình giải tiếp bài 2
Bài 2
a) \(2x\left(x^2-3\right)=2x^3-6x\)
b) \(x\left(x^2-2x+5\right)=x^3-2x^2+5x\)
c) \(\left(x+2y\right)\left(x+2y^2-5xy\right)\)
\(=x\left(x+2y^2-5xy\right)+2y\left(x+2y^2-5xy\right)\)
\(=x^2+2xy^2-5x^2y+2xy+4y^3-10xy^2\)
\(=4y^3+x^2-8xy^2-5x^2y+2xy\)
d)Tương tự bài c)
\(A=x^2-6x+10=x^2-2.3x+3^2+1=\left(x-3\right)^2+1\)
Ta có: \(\left(x-3\right)^2\ge0\) nên \(\left(x-3\right)^2+1\ge1\)
Vậy \(A_{min}=1\)(Dấu "="\(\Leftrightarrow x=3\))
a) \(\left(x+1\right)^3-x^2\left(x+3\right)=2\)
\(\Leftrightarrow\left(x^3+3x^2+3x+1\right)-\left(x^3+3x^2\right)=2\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3-3x^2=2\)
\(\Leftrightarrow3x+1=2\)
\(\Leftrightarrow3x=1\)
\(\Leftrightarrow x=\frac{1}{3}\)
a) \(\left(2x+3\right)^2-3\left(x-4\right)\left(x+4\right)=\left(x-2\right)^2+1\)
\(\Leftrightarrow4x^2+12x+9-3\left(x^2-16\right)=x^2-4x+4+1\)
\(\Leftrightarrow4x^2+12x+9-3x^2+48=x^2-4x+5\)
\(\Leftrightarrow x^2+12x+57=x^2-4x+5\)
\(\Leftrightarrow16x+52=0\)
\(\Leftrightarrow x=-\frac{13}{4}\)
b) \(\left(3x-2\right)\left(9x^2+6x+4\right)-\left(3x-1\right)\left(9x^2-3x+1\right)=x-4\)
\(\Leftrightarrow\)Xem lại đề !
c) \(x\left(x-1\right)-\left(x-3\right)\left(x+4\right)=5x\)
\(\Leftrightarrow x^2-x-x^2-x+12=5x\)
\(\Leftrightarrow-2x+12=5x\)
\(\Leftrightarrow7x-12=0\)
\(\Leftrightarrow x=\frac{12}{7}\)
d) \(\left(2x+1\right)\left(2x-1\right)=4x\left(x-7\right)-3x\)
\(\Leftrightarrow4x^2-1=4x^2-28x-3x\)
\(\Leftrightarrow28x+3x-1=0\)
\(\Leftrightarrow31x-1=0\)
\(\Leftrightarrow x=\frac{1}{31}\)
Bài 1:
\(P=3x^2+x-1\)
\(=3\left(x^2+\frac{1}{3}x-\frac{1}{3}\right)\)
\(=3\left(x^2+2x.\frac{1}{6}+\frac{1}{36}-\frac{13}{36}\right)\)
\(=3\left(x+\frac{1}{6}\right)^2-\frac{13}{12}\ge\frac{-13}{12}\)\(\forall x\)
Dấu '' = '' xảy ra khi: \(\left(x+\frac{1}{6}\right)^2=0\Rightarrow x=\frac{-1}{6}\)
Vậy \(MinP=\frac{-13}{12}\) khi \(x=\frac{-1}{6}\)
Bài 2:
a) Không có điều kiện
b) Nghiệm vô tỉ
Bạn xem lại đề hai phần này nhé.
c) \(\left(x-2\right)^3-x^3+6x^2=14\)
\(\Rightarrow x^3-6x^2+12x-8-x^3+6x^2-14=0\)
\(\Rightarrow\left(x^3-x^3\right)+\left(-6x^2+6x^2\right)+12x+\left(-8-14\right)=0\)
\(\Rightarrow12x-22=0\)
\(\Rightarrow x=\frac{11}{6}\)
d) \(8x^2+30x+7=0\)
\(\Rightarrow8x^2+28x+2x+7=0\)
\(\Rightarrow\left(8x^2+28x\right)+\left(2x+7\right)=0\)
\(\Rightarrow4x\left(2x+7\right)+\left(2x+7\right)=0\)
\(\Rightarrow\left(4x+1\right)\left(2x+7\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4x+1=0\\2x+7=0\end{cases}}\Rightarrow\orbr{\begin{cases}4x=-1\\2x=-7\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-1}{4}\\x=\frac{-7}{2}\end{cases}}\)