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LD
18 tháng 9 2018
Bài 1:
a) \(9\left(4x+3\right)^2=16\left(3x-5\right)^2\)
\(114x^2+216x+81=114x^2-480x+400\)
\(144x^2+216x=144x^2-480x+400-81\)
\(114x^2+216=114x^2-480x+319\)
\(696x=319\)
\(\Rightarrow x=\frac{11}{24}\)
b) \(\left(x^3-x^2\right)^2-4x^2+8x-4=0\)
\(\left(x-1\right)^2\left(x^2+2\right)\left(x+\sqrt{2}\right)\left(x-\sqrt{2}\right)=0\)
\(\Rightarrow x=1\)
c) \(x^5+x^4+x^3+x^2+x+1=0\)
\(\left(x+1\right)\left(x^2+x+1\right)\left(x^2-x+1\right)=0\)
\(\Rightarrow x=-1\)
Bài 2:
a) \(5x^3-7x^2-15x+21=0\)
\(\left(5x-7\right)\left(x+\sqrt{3}\right)\left(x-\sqrt{3}\right)=0\)
\(\Rightarrow x=\frac{7}{5}\)
b) \(\left(x-3\right)^2=4x^2-20x+25\)
\(x^2-6x+9-25=4x^2-20x+25\)
\(x^2-6x+9=4x^2-20x+25-25\)
\(x^2-6x-16=4x^2-20x\)
\(x^2+14x-16=4x^2-4x^2\)
\(-3x^2+14x-16=0\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=\frac{8}{3}\end{cases}}\)
c) \(\left(x-1\right)^2-5=\left(x+2\right)\left(x-2\right)-x\left(x-1\right)\)
\(x^2-2x=x-4\)
\(x^2-2x=x-4+4\)
\(x^2-2x=x-x\)
\(x^2-3x=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
d) \(\left(2x-3\right)^3-\left(2x+3\right)\left(4x^2-1\right)=-24\)
\(-48x^2+56x-24=-24\)
\(-48x^2+56x=-24+24\)
\(-48x^2+56=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{7}{6}\end{cases}}\)
mình ko chắc
TN
28 tháng 7 2016
a)(2x-3)2=(x+5)2
=>4x2-12x+9=x2+10x+25
=>3x2-22x-16=0
=>3x2+2x-24x-16=0
=>x(3x+2)-8(3x+2)=0
=>(x-8)(3x+2)=0
=>x=8 hoặc x=-2/3
b)X2.(x-1)-4x2+8x-4=0
=>x2(x-1)-4x2+4x+4x-4=0
=>x2(x-1)-4x(x-1)-4(x-1)=0
=>x2(x-1)-(4x-4)(x-1)=0
=>(x2-4x+4)(x-1)=0
=>(x-2)2(x-1)=0
=>x=2 hoặc x=1
c) 4x2- 25 - (2x- 5) . ( 2x+7)=0
=>4x2-25-(4x2+14x-10x-35)=0
=>4x2-25-4x2-14x+10x+35=0
=>-4x+10=0
=>-4x=-10 <=>x=5/2
d) x3+27+(x+3).(x-9)=0
=>x3+33+(x+3)(x-9)=0
=>(x+3)(x2-3x+9)+(x+3)(x-9)=0
=>(x2-3x+9+x-9)(x+3)=0
=>(x2-2x)(x+3)=0
=>x(x-2)(x+3)=0
=>x=0 hoặc x=2 hoặc x=-3
e) (x-2).(x+5)- x2+4=0
=>(x-2)(x+5)-(x-2)(x+2)=0
=>(x-2)(x+5-x-2)=0
=>3(x-2)=0 <=>x=2
28 tháng 7 2016
Sau khi khai triển hằng đẳng thức và thực hiện chuyển vế bạn sẽ đk kết quả như này!(\(\left(2x-3\right)^2=\left(x+5\right)^2=3x^2-22x-14\)
KN
12 tháng 7 2019
g) \(\left(2x-1\right)^2-\left(2x+4\right)^2=0\)
\(\Leftrightarrow\left(2x-1+2x+4\right)\left(2x-1-2x-4\right)=0\)
\(\Leftrightarrow-5\left(4x+3\right)=0\)
\(\Leftrightarrow4x+3=0\)
\(\Leftrightarrow4x=-3\)
\(\Leftrightarrow x=\frac{-3}{4}\)
Vậy tập nghiệm của pt là \(S=\left\{\frac{-3}{4}\right\}\)
KN
12 tháng 7 2019
h) \(\left(2x-3\right)\left(3x+1\right)-x\left(6x+10\right)=30\)
\(\Leftrightarrow3x\left(2x-3\right)+\left(2x-3\right)-6x^2-10x=30\)
\(\Leftrightarrow6x^2-9x+2x-3-6x^2-10x=30\)
\(\Leftrightarrow-9x+2x-3-10x=30\)
\(\Leftrightarrow-17x-3=30\)
\(\Leftrightarrow-17x=33\)
\(\Leftrightarrow x=\frac{-33}{17}\)
Vậy tập nghiệm của pt là \(S=\left\{\frac{-33}{17}\right\}\)
R
9 tháng 7 2019
Bài 1:tìm x ,biết:
a) (2x - 1)(3x + 2) - 6x(x + 1) = 0
\(\Leftrightarrow6x^2+x-2-6x^2-6x=0\)
\(\Leftrightarrow-5x=2\)
\(\Leftrightarrow x=\frac{-2}{5}\)
b) \(\left(4x-1\right)^2-\left(2x+1\right)\left(8x-3\right)=0\)
\(\Leftrightarrow16x^2-8x+1-16x^2-2x+3=0\)
\(\Leftrightarrow-10x=-4\)
\(\Leftrightarrow x=\frac{2}{5}\)
c) \(4x^2-1=2\left(2x+1\right)\)
\(\Leftrightarrow\left(2x+1\right)\left(2x-1\right)-2\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{3}{2}\end{cases}}\)
2a) \(4x^2-9y^2-6y-1=4x^2-\left(3y+1\right)^2\)
\(=\left(2x-3y-1\right)\left(2x+3y+1\right)\)
b) \(4x^2-1-2x\left(2x-1\right)=\left(2x-1\right)\left(2x+1\right)-2x\left(2x-1\right)\)
\(=1.\left(2x-1\right)\)
c) \(x^2-8x-4y^2+16=\left(x-4\right)^2-4y^2\)
\(=\left(x-4-2y\right)\left(x-4+2y\right)\)
d) \(9x^2-12x-y^2+4=\left(3x-2\right)^2-y^2\)
\(=\left(3x-2-y\right)\left(3x-2+y\right)\)
e) \(4x^2+10x-5=4x^2+2.2.\frac{5}{2}x+\frac{25}{4}-\frac{25}{4}-5\)
\(=\left(2x+\frac{5}{2}\right)^2-\frac{45}{4}\)
\(=\left(2x+\frac{5+3\sqrt{5}}{2}\right)\left(2x+\frac{5-3\sqrt{5}}{2}\right)\)
15 tháng 11 2019
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
KN
15 tháng 11 2019
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)
\(=\left(7x+6-5+6x\right)^2\)
\(=\left(13x+1\right)^2\)
HL
18 tháng 7 2016
a) (2x+3)(4x2-6x+9)-2(4x3-1)+(8x-1)=15
<=>8x3+27-8x3+2+8x-1=15
<=>8x+28=15
<=>8x=-13
<=>x=-13/8
b) (x+3)3-(x+9)(x2+27)-(5x-216) = 3x-4
<=>x3+9x2+27x+27-x3-27x-9x2-243-5x+216=3x-4
<=>-5x=3x-4
<=>8x=4
<=>x=1/2
3 tháng 8 2019
\(x\left(2x-7\right)-4x+14=0\Leftrightarrow\left(x-2\right)\left(2x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{7}{2}\end{matrix}\right.\)
\(x^2\left(x-1\right)-4\left(x-1\right)=\left(x^2-4\right)\left(x-1\right)=\left(x-2\right)\left(x+2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=1\end{matrix}\right.\)
\(x^4-x^3-x^2+x=x\left(x^3+1\right)-x^2\left(x+1\right)=x\left(x+1\right)\left(x^2-x+1-x^2\right)=x\left(x+1\right)\left(1-x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\\1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm1\end{matrix}\right.\)
3 tháng 8 2019
a) \(x\left(2x-7\right)-4x+14-0\Leftrightarrow2x^2-11x+14=0\Leftrightarrow2x^2-4x-7x+14=0\Leftrightarrow2x\left(x-2\right)-7\left(x-2\right)=0\Leftrightarrow\left(2x-7\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3,5\\x=2\end{matrix}\right.\)
b) \(x^2\left(x-1\right)-4x+4=0\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-2\end{matrix}\right.\)
c) \(x+x^2-x^3-x^4=0\Leftrightarrow x\left(x^3+x^2-x-1\right)=0\Leftrightarrow x\left[x\left(x^2-1\right)+\left(x^2-1\right)\right]=0\Leftrightarrow x\left(x+1\right)\left(x^2-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
d) \(2x^3+3x^2+2x+3=0\Leftrightarrow x^2\left(2x+3\right)+2x+3=0\Leftrightarrow\left(x^2+1\right)\left(2x+3\right)=0\Leftrightarrow x=-1,5\left(x^2+1>0\forall x\right)\)
e) \(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\Leftrightarrow\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)=0\Leftrightarrow\left(2x-5\right)\left(2x+5-2x-7\right)=0\Leftrightarrow2x-5=0\Leftrightarrow x=2,5\)
g) \(x^3+27+\left(x+3\right)\left(x-9\right)=0\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\Leftrightarrow x\left(x+3\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=2\end{matrix}\right.\)
\(a.\left(x+1\right)\left(x^2-x+1\right)-x\left(x^2-5\right)=71\)
\(\Leftrightarrow x^3+1-x^3+5x=71\)
\(\Leftrightarrow5x=71-1\)
\(\Leftrightarrow5x=70\)
\(\Leftrightarrow x=70:5=14\)
\(b.\left(2x-3\right)^3-8x\left(x-1\right)^2+4x\left(4x+1\right)+27=0\)
\(\Leftrightarrow8x^3-12x^2+18x-27-8x\left(x^2-2x+1\right)+16x^2+4x+27=0\)
\(\Leftrightarrow8x^3-12x^2+18x-27-8x^3+16x^2-8x+16x^2+4x+27=0\)
\(\Leftrightarrow20x^2+14x=0\)
\(\Leftrightarrow x\left(20x+14\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\20x+14=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{7}{10}\end{cases}}}\)
a) ta có: (x+1)(x^2 -x+1) -x(x^2 -5)=71
<=>x^3 +1 -x^3 +5x=71
<=>5x=70
<=>x=14
b) ta có:(2x-3)^3 -8x(x-1)^2 +4x(4x+1)+27=0
<=>[ (2x-3)^3 +27)] - [ 8x(x-1)^2 -4x(4x+1)]=0
<=> (2x-3+3)[ (2x-3)^2 - (2x-3).3 +3^2] - 2x [ 4(x^2 -2x +1) -2(4x+1)]=0
<=>2x( 4.x^2 - 12x +9 - 6x +9 +9) - 2x( 4.x^2 -8x+4 -8x -2)=0
<=>2x(4.x^2 -18x +27) - 2x(4.x^2 -16x +2)=0
<=>2x(4.x^2 -18x+27 -4.x^2 +16x-2)=0
<=>2x(25-2x)=0
<=>x=0 hoặc 25-2x=0 <=> x=0 hoặc x=25/2