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24 tháng 9 2017
Bài1:
\(a,\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=22\\ \Leftrightarrow\left(2x+3\right)^2-4x^2+1=22\\ \Leftrightarrow\left(2x+3-2x\right)\left(2x+3+2x\right)=21\\ \Leftrightarrow3\left(4x+3\right)=21\\ \Leftrightarrow4x+3=7\\ \Leftrightarrow4x=4\\ \Leftrightarrow x=1\\ Vậy....\\ b,\left(2x-1\right)^3-4x^2\left(2x-3\right)=5\\ \Leftrightarrow8x^3-12x^2+6x-1-8x^3+12x^2=5\\ \Leftrightarrow6x=6\\ \Leftrightarrow x=1\\ Vậy...\)
Các câu sau cũng như thế
Bài2:
\(A=x^2+20x+9\\ =\left(x^2+20x+100\right)-91\\ =\left(x+10\right)^2-91\)
Với mọi x thì \(\left(x+10\right)^2\ge0\\ \Rightarrow\left(x+10\right)^2-91\ge-91\)
Hay \(A\ge-91\)
Để A=-91 thì
\(\left(x+10\right)^2=0\\ \Leftrightarrow x+10=0\\ \Leftrightarrow x=-10\)
Vậy...
\(B=4x^2+5x+7\\ =\left(4x^2+5x+\dfrac{25}{16}\right)+5,4375\\ =\left(2x+\dfrac{5}{4}\right)^2+5,4375\)
Với mọi x;y thì \(\left(2x+\dfrac{5}{4}\right)^2+5,4375\ge5,4375\)
Hay \(A\ge5,4375\)
Để \(A=5,4375\) thì \(\left(2x+\dfrac{5}{4}\right)^2=0\\ \Leftrightarrow2x+\dfrac{5}{4}=0\\ \Leftrightarrow x=\dfrac{-5}{8}\)
Vậy....
22 tháng 5 2022
a: Đặt \(C=3\left(x-1\right)^2-\left(x+1\right)^2+2\left(x-3\right)\left(x+3\right)-\left(2x+3\right)^2-\left(5-20x\right)\)
\(D=5x\left(x-7\right)\left(x+7\right)-x\left(2x-1\right)^2-\left(x^3+4x^2-246x\right)-175\)
Do đó: A=C+D
\(C=3\left(x-1\right)^2-\left(x+1\right)^2+2\left(x-3\right)\left(x+3\right)-\left(2x+3\right)^2-\left(5-20x\right)\)
\(=3x^2-6x+3-x^2-2x-1+2x^2-18-\left(4x^2+12x+9\right)-5+20x\)
\(=4x^2-8x-16-4x^2-12x-9-5+20x\)
\(=-30\)
\(D=5x\left(x-7\right)\left(x+7\right)-x\left(2x-1\right)^2-\left(x^3+4x^2-246x\right)-175\)
\(=5x\left(x^2-49\right)-x\left(4x^2-4x+1\right)-x^3-4x^2+246x-175\)
\(=5x^3-245x-4x^3+4x^2-x-x^3-4x^2+246x-175\)
=-175
A=C+D=-30-175=-205
b: Đặt \(E=-2x\left(3x+2\right)^2+\left(4x+1\right)^2+2\left(x^3+8x^2+3x-2\right)-\left(5-x\right)\)
\(F=\left(5x-2\right)^2-\left(6x+1\right)^2+11\left(x-2\right)\left(x+2\right)-16\left(3-2x\right)\)
Do đó: B=E+F
\(E=-2x\left(3x+2\right)^2+\left(4x+1\right)^2+2\left(x^3+8x^2+3x-2\right)-\left(5-x\right)\)
\(=-2x\left(9x^2+12x+4\right)+16x^2+8x+1+2x^3+16x^2+6x-4-5+x\)
\(=-18x^3-24x^2-8x+32x^2+14x+1-5+x\)
\(=-18x^3+8x^2+7x-4\)
\(F=\left(5x-2\right)^2-\left(6x+1\right)^2+11\left(x-2\right)\left(x+2\right)-16\left(3-2x\right)\)
\(=25x^2-20x+4-36x^2-12x-1+11x^2-44-48+32x\)
\(=-95\)
\(B=-18x^3+8x^2+7x-99\)
19 tháng 5 2022
Bài 1:
a: \(A=3\left(x^2-2x+1\right)-\left(x^2+2x+1\right)+2\left(x^2-9\right)-\left(4x^2+12x+9\right)-5+20x\)
\(=3x^2-6x+3-x^2-2x-1+2x^2-18-\left(4x^2+12x+9\right)-5+20x\)
\(=4x^2-8x-16-5+20x-4x^2-12x-9\)
\(=-30\)
b: \(B=5x\left(x^2-49\right)-x\left(4x^2-4x+1\right)-\left(x^3+4x^2-246x\right)-175\)
\(=5x^3-245x-4x^3+4x^2-x-x^3-4x^2+246x-175\)
\(=-175\)
d: \(D=25x^2-20x+4-36x^2-12x-1+11\left(x^2-4\right)-48+32x\)
\(=-11x^2-32x+3-48+32x+11x^2-44\)
=-89
27 tháng 7 2022
a: \(\Leftrightarrow8x^3-4x+27=8x^3+8x^2+12x^2+12x+18x+18\)
\(\Leftrightarrow8x^3+20x^2+30x+18=8x^3-4x+27\)
\(\Leftrightarrow20x^2+34x-9=0\)
hay \(x\in\left\{\dfrac{-17+\sqrt{469}}{20};\dfrac{-17-\sqrt{469}}{20}\right\}\)
b: \(\Leftrightarrow20x^2-16x-1=10x^2-2x+5x-1=10x^2+3x-1\)
\(\Leftrightarrow10x^2-19x=0\)
=>x=0 hoặc x=19/10
3 tháng 8 2017
a. 5x.(12x+7)-3x.(20x-5)=-150
x=-3
b. ( 2x-1).(3-x)+(x+4).(2x-5)=20
x=43/10
c. 9x2-1+(3x-1)2=0
x=1/3
d. 3x.(x-2)-(3x+2).(x-1)=7
x=-5/2
e. (2x-1)2-(2x+5).(2x-5)=20
x=3/2
f. 4x2-5=4
x=3/2
~~~~~~~~~~~ai đi ngang qua nhớ để lại k ~~~~~~~~~~~~~
~~~~~~~~~~~~ Chúc bạn sớm kiếm được nhiều điểm hỏi đáp ~~~~~~~~~~~~~~~~~~~
NH
1
24 tháng 5 2022
b: ta có: \(x^2-5x=-6\)
\(\Leftrightarrow x^2-5x+6=0\)
=>(x-2)(x-3)=0
=>x=2 hoặc x=3
c: Sửa đề: \(\left(2x-1\right)^2-\left(3x+5\right)^2=0\)
\(\Leftrightarrow\left(2x-1-3x-5\right)\left(2x-1+3x+5\right)=0\)
\(\Leftrightarrow\left(-x-6\right)\left(5x+4\right)=0\)
=>x=-6 hoặc x=-4/5
d: ta có: \(4x^2-20x+25=0\)
\(\Leftrightarrow\left(2x-5\right)^2=0\)
=>2x-5=0
hay x=5/2
e: \(\Leftrightarrow\left(3x-1-x+2\right)\left(3x-1+x-2\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(4x-3\right)=0\)
hay \(x\in\left\{-\dfrac{1}{2};\dfrac{3}{4}\right\}\)
CM
4 tháng 12 2018
a) \(2x\left(x-3\right)+5\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(2x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\2x=-5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)
b) \(\left(x^2-4\right)-\left(x-2\right)\left(3-2x\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x+2-3+2x\right)=0\)
\(\Rightarrow\left(x-2\right)\left(3x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\3x=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
c) \(\left(2x+5\right)^2=\left(x+2\right)^2\)
\(\Rightarrow\left(2x+5\right)^2-\left(x+2\right)^2=0\)
\(\Rightarrow\left(2x+5-x-2\right)\left(2x+5+x+2\right)=0\)
\(\Rightarrow\left(x+3\right)\left(3x+7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\3x+7=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\3x=-7\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{7}{3}\end{matrix}\right.\)
d) \(x^2-5x+6=0\)
\(\Rightarrow x^2-2x-3x+6=0\)
\(\Rightarrow x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
e) \(2x^3+6x^2=x^2+3x\)
\(\Rightarrow2x^3+6x^2-x^2-3x=0\)
\(\Rightarrow2x^3+5x^2-3x=0\)
\(\Rightarrow x\left(2x^2+5x-3\right)=0\)
\(\Rightarrow2x^2+5x-3=0\)
\(\Rightarrow2x^2-6x+x-3=0\)
\(\Rightarrow2x\left(x-3\right)+\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(2x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\2x+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{2}\end{matrix}\right.\)
f) \(\left(x^2-1\right)\left(x+2\right)-\left(x-2\right)\left(x^2+2x+4\right)-2x^2\)
\(\Rightarrow\left(x^2-1\right)\left(x+2\right)-\left(x^3-8\right)-2x^2=0\)
\(\Rightarrow x^3+2x^2-x+2-x^3+8-2x^2=0\)
\(\Rightarrow-x+10=0\)
\(\Rightarrow x=10\)
22 tháng 5 2022
a: Đặt \(C=3\left(x-1\right)^2-\left(x+1\right)^2+2\left(x-3\right)\left(x+3\right)-\left(2x+3\right)^2-\left(5-20x\right)\)
\(D=5x\left(x-7\right)\left(x+7\right)-x\left(2x-1\right)^2-\left(x^3+4x^2-246x\right)-175\)
Do đó: A=C+D
\(C=3\left(x-1\right)^2-\left(x+1\right)^2+2\left(x-3\right)\left(x+3\right)-\left(2x+3\right)^2-\left(5-20x\right)\)
\(=3x^2-6x+3-x^2-2x-1+2x^2-18-\left(4x^2+12x+9\right)-5+20x\)
\(=4x^2-8x-16-4x^2-12x-9-5+20x\)
\(=-30\)
\(D=5x\left(x-7\right)\left(x+7\right)-x\left(2x-1\right)^2-\left(x^3+4x^2-246x\right)-175\)
\(=5x\left(x^2-49\right)-x\left(4x^2-4x+1\right)-x^3-4x^2+246x-175\)
\(=5x^3-245x-4x^3+4x^2-x-x^3-4x^2+246x-175\)
=-175
A=C+D=-30-175=-205
b: Đặt \(E=-2x\left(3x+2\right)^2+\left(4x+1\right)^2+2\left(x^3+8x^2+3x-2\right)-\left(5-x\right)\)
\(F=\left(5x-2\right)^2-\left(6x+1\right)^2+11\left(x-2\right)\left(x+2\right)-16\left(3-2x\right)\)
Do đó: B=E+F
\(E=-2x\left(3x+2\right)^2+\left(4x+1\right)^2+2\left(x^3+8x^2+3x-2\right)-\left(5-x\right)\)
\(=-2x\left(9x^2+12x+4\right)+16x^2+8x+1+2x^3+16x^2+6x-4-5+x\)
\(=-18x^3-24x^2-8x+32x^2+14x+1-5+x\)
\(=-18x^3+8x^2+7x-4\)
\(F=\left(5x-2\right)^2-\left(6x+1\right)^2+11\left(x-2\right)\left(x+2\right)-16\left(3-2x\right)\)
\(=25x^2-20x+4-36x^2-12x-1+11x^2-44-48+32x\)
\(=-95\)
\(B=-18x^3+8x^2+7x-99\)
2 tháng 1 2017
Bài 1:
\(\left(2x-1\right)^2-4x^2\left(2x-3\right)=5\)
\(\Rightarrow4x^2-4x+1-8x^3+12x^2-5=0\)
\(\Rightarrow-8x^3+16x^2-4x-4=0\)
\(\Rightarrow-4\left(2x^3-4x^2+x+1\right)=0\)
\(\Rightarrow2x^3-4x^2+x+1=0\)
\(\Rightarrow2x^3-2x^2-x-2x^2+2x+1=0\)
\(\Rightarrow x\left(2x^2-2x-1\right)-\left(2x^2-2x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(2x^2-2x-1\right)=0\)
\(\Rightarrow\left[\begin{matrix}x-1=0\\2x^2-2x-1=0\end{matrix}\right.\)\(\Rightarrow\left[\begin{matrix}x=1\\\Delta_{2x^2-2x-1}=\left(-2\right)^2-\left(-4\left(2.1\right)\right)=12\end{matrix}\right.\)
\(\Rightarrow\left[\begin{matrix}x=1\\x_{1,2}=\frac{2\pm\sqrt{12}}{4}\end{matrix}\right.\)
2 tháng 1 2017
Bài 2:
a)\(x^2+6x-y^2+9\)
\(=\left(x^2+6x+9\right)-y^2\)
\(=\left(x+3\right)^2-y^2\)
\(=\left(x+3-y\right)\left(x+3+y\right)\)
b)\(2x^2-3xy-2y^2+5x+5y-3\)
\(=2x^2+xy-x-4xy-2y^2+2y+6x+3y-3\)
\(=x\left(2x+y-1\right)-2y\left(2x+y-1\right)+3\left(2x+y-1\right)\)
\(=\left(x-2y+3\right)\left(2x+y-1\right)\)
\(\left(2x-1\right)^3-2x\left(2x+1\right)^2=5x-20x^2\)
\(\Rightarrow8x^3-12x^2+6x-1-2x\left(4x^2+4x+1\right)-5x+20x^2=0\)
\(\Rightarrow8x^3-12x^2+6x-1-8x^3-8x^2-2x-5x+20x^2=0\)
\(\Rightarrow-x-1=0\)
\(\Rightarrow x=-1\)