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Bài 2:
Ta có: \(P=3x\left(\dfrac{2}{3}x^2-3x^4\right)+9x^2\left(x^3-1\right)+x^2\left(-2x+9\right)-12\)
\(=2x^3-9x^5+9x^5-9x^2-2x^3+9x^2-12\)
=-12
Bài 1:
a: Ta có: \(x\left(x^2+2\right)+2x\left(1-\dfrac{1}{2}x^2\right)=4\)
\(\Leftrightarrow x^3+2x+2x-x^3=4\)
hay x=1
b: Ta có: \(4x^2\left(x-1\right)+x\left(x^2+4x\right)=40\)
\(\Leftrightarrow4x^3-4x^2+x^3+4x^2=40\)
\(\Leftrightarrow5x^3=40\)
hay x=2
c: Ta có: \(3x\left(x-2\right)-3\left(x^2-3\right)=8\)
\(\Leftrightarrow3x^2-6x-3x^2+9=8\)
\(\Leftrightarrow-6x=-1\)
hay \(x=\dfrac{1}{6}\)
b: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(x^2-2x-3\right)=0\)
=>(7x+10)(x-3)=0
hay \(x\in\left\{-\dfrac{10}{7};3\right\}\)
d: \(\Leftrightarrow\dfrac{13}{2x^2+7x-6x-21}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{\left(2x+7\right)}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow26x+91+x^2-9-12x-14=0\)
\(\Leftrightarrow x^2+14x+68=0\)
hay \(x\in\varnothing\)
\(a,\Leftrightarrow\left(3x-7\right)\left(3x+7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-\dfrac{7}{3}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-1-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\\ c,\Leftrightarrow4x^2-7x-2-4x^2+4x+3=7\\ \Leftrightarrow-3x=6\Leftrightarrow x=-2\\ d,\Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25=0\\ \Leftrightarrow4x=-26\Leftrightarrow x=-\dfrac{13}{2}\\ e,\Leftrightarrow x^3+27-x^3+x-27=0\\ \Leftrightarrow x=0\\ f,\Leftrightarrow\left(4x-3\right)\left(4x-3+3x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{7}\end{matrix}\right.\)
a) 9x2-49=0
(3x)2-72=0
<=> (3x-7)(3x+7)=0
th1: 3x-7=0
<=>3x=7
<=>x=\(\dfrac{7}{3}\)
th2: 3x+7=0
<=>3x=-7
<=>x=\(-\dfrac{7}{3}\)
a) \(9x^2-49=0\)
\(\Rightarrow\left(3x-7\right)\left(3x+7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3x+7\\3x-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\frac{7}{3}\\x=\frac{7}{3}\end{matrix}\right.\)
Mấy í sau đến chịu k dịch đc
a: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\cdot\left(x^2-2x-3\right)=0\)
=>(7x+10)(x-3)=0
=>x=3 hoặc x=-10/7
b: \(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow13\left(x+3\right)+x^2-9-12x-42=0\)
\(\Leftrightarrow x^2-12x-51+13x+39=0\)
\(\Leftrightarrow x^2+x-12=0\)
=>(x+4)(x-3)=0
=>x=-4
a ) \(9x^2-49=9\)
\(\Leftrightarrow9x^2=58\)
\(\Leftrightarrow x^2=29\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=29\\x=-29\end{array}\right.\)
Vậy ......................
b ) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-1\right)\left(x+1\right)-27=0\)
\(\Leftrightarrow\left(x^3+3^3\right)-x.\left(x^2-1^2\right)-27=0\)
\(\Leftrightarrow x^3+27-x^3+x-27=0\)
\(\Leftrightarrow x=0\)
c ) \(\left(x-1\right)\left(x+2\right)-x-2=0\)
\(\Leftrightarrow x^2+2x-x-2-x-2=0\)
\(\Leftrightarrow x^2-4=0\)
\(\Leftrightarrow x^2=4\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-2\end{array}\right.\)
Vây .....................
Giải như sau.
(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y
⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn !
\(\left(x+6\right)\left(2x+1\right)=0\)
<=> \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)
Vậy....
hk tốt
^^
a/ \(x^2+y^2=x^2+y^2+2xy-2xy =\left(x+y\right)^2-2xy\)
b/ mình không chắc nữa
bài 3
a/ \(9x^2-49=0 \Leftrightarrow x^2=\frac{49}{9} \Leftrightarrow\orbr{\begin{cases}x=\frac{7}{3}\\x=-\frac{7}{3}\end{cases}}\)
b/ \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x+1\right)\left(x-1\right)-27=0 \Leftrightarrow x^3+27-x\left(x^2-1\right)-27=0\)
\(\Leftrightarrow x^3-x^3+x=0\Leftrightarrow x=0\)
c/\(\left(x-1\right)\left(x+2\right)-x-2=0 \Leftrightarrow \left(x-1\right)\left(x+2\right)-\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)^2=0\Leftrightarrow\orbr{\begin{cases}x+2=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=1\end{cases}}}\)
d/ \(x\left(3x+2\right)+\left(x+1\right)^2-\left(2x-5\right)\left(2x+5\right)=0\)
\(\Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25=0\)
\(\Leftrightarrow4x+25=0 \Leftrightarrow x=\frac{-25}{4}\)
e/ mình lười qá ko viết đề đâu
\(\Leftrightarrow4x^2-7x-2-4x^2+4x+3=7\)
\(\Leftrightarrow-3x+1=7 \Leftrightarrow x=-2\)
có gì sai bn sửa lại nha
b: =>(x-8)(x+5)=0
=>x=8 hoặc x=-5
b) (x-8)(x+5)=0
x=8