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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 .....................
A) ( x + 2)2 - ( x -2)(x +2) = 0
<=>(x+2)[(x+2)-(x-2)]=0
<=>(x+2)*4=0
<=>x+2=0 <=>x=-2
B) (3x + 2) + (x + 1)2 - (2x - 5)(2x + 5) = -12
<=>3x+2+x2+2x+1-4x2+25+12=0
<=>-3x2+5x+40
\(\Delta=5^2-\left(-4\left(3.40\right)\right)=505\)
\(\Rightarrow x_{1,2}=\frac{-5\pm\sqrt{505}}{6}\)
a) \(\Leftrightarrow16x^2-\left(16x^2-40x+25\right)=15\)
\(\Leftrightarrow16^2-16x^2+40x+25-15=0\)
\(\Leftrightarrow40x+10=0\)
\(\Leftrightarrow x=-\frac{10}{40}=-\frac{1}{4}\)
b)\(\Leftrightarrow4x^2+12x+9-4\left(x^2-1\right)=49\)
\(\Leftrightarrow4x^2+12x+9-4x^2+4-49=0\)
\(\Leftrightarrow12x-36=0\)
\(\Leftrightarrow x=\frac{36}{12}=3\)
c) \(\Leftrightarrow9x^2-6x+1-\left(9x^2-12x+4\right)=0\)
\(\Leftrightarrow9x^2-6x+1-9x^2+12x-4=0\)
\(\Leftrightarrow6x-3=0\)
\(\Leftrightarrow x=\frac{3}{6}=\frac{1}{2}\)
nha Nhấp Đúng nha . Chúc bạn học tốt!!!!Cảm ơn !
a) (3x-5)2 - (x+1)2 =0
\(\Leftrightarrow\left(3x-5+x+1\right)\left[\left(3x-5\right)-\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(4x-4\right)\left(2x-6\right)=0\)
\(\Leftrightarrow8\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x-3=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=3\end{array}\right.\)
b) 4x3 - 36x =0
\(\Leftrightarrow4x\left(x^2-9\right)=0\)
\(\Leftrightarrow4x\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}4x=0\\x-3=0\\x+3=0\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=3\\x=-3\end{array}\right.\)
a. \(\left(3x-5\right)^2-\left(x+1\right)^2=0\Leftrightarrow\left(3x-5+x+1\right)\left(3x-5-x-1\right)=0\Leftrightarrow\left(4x-4\right)\left(2x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}4x-4=0\\2x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\)
Vậy ...
b. \(\left(5x-4\right)^2-49x^2=0\Leftrightarrow\left(5x-4\right)^2-\left(7x\right)^2=0\Leftrightarrow\left(5x-4-7x\right)\left(5x-4+7x\right)=0\Leftrightarrow\left(-2x-4\right)\left(12x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}-2x-4=0\\12x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy ...
c. \(4x^3-36x=0\Leftrightarrow4x\left(x^2-9\right)=0\Leftrightarrow4x\left(x-3\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}4x=0\\x-3=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
Vậy ...
d. \(\left(2x+3\right)\left(x-1\right)+\left(2x-3\right)\left(1-x\right)=0\Leftrightarrow\left(2x+3\right)\left(x-1\right)-\left(2x-3\right)\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(2x+3-2x+3\right)=0\Leftrightarrow6\left(x-1\right)=0\Leftrightarrow x-1=0\Leftrightarrow x=1\)
Vậy ...
a. 3.(x-2)+2.(x-3)=13
x=5
b. (x+1).(2-x)-(3x+5).(x+2)=-4x2+1
x=-9/10
c.x.(5-2x)+2x.(x-1)=13
x=13/3
d. (2x+3)2-(x-1)2=0
x=-2/3
e. x2.(3x-2)-8+12=0
x vô ngiệm
f x2+x=0
x=-1
g. x3-5x=0
x=0
~~~~~~~~~~~ai đi ngang qua nhớ để lại k ~~~~~~~~~~~~~
~~~~~~~~~~~~ Chúc bạn sớm kiếm được nhiều điểm hỏi đáp ~~~~~~~~~~~~~~~~~~~
a) \(3\left(x-2\right)+2\left(x-3\right)=1\)\(3\)
\(3x-6+2x-6=13\)
\(5x=13+6+6\)
\(5x=25\)
\(x=25\)
c) \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(5x-2x^2+2x^2-2x=13\)
\(3x=13\)
\(x=\frac{13}{3}\)
d) \(\left(2x+3\right)^2-\left(x-1\right)^2=0\)
\(\left(2x+3-x+1\right)\left(2x+3+x-1\right)=0\)
\(\left(x+4\right)\left(3x+2\right)=0\)
\(\orbr{\begin{cases}x+4=0\\3x+2=0\end{cases}}=>\orbr{\begin{cases}x=-4\\x=\frac{-2}{3}\end{cases}}\)
f) \(x^2+x=0\)
\(x\left(x+1\right)=0\)
\(=>\orbr{\begin{cases}x=0\\x+1=0\end{cases}=>\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)
g) \(x^3-5x=0\)
\(x^2\left(x-5\right)=0\)
\(=>\orbr{\begin{cases}x^2=0\\x-5=0\end{cases}}\)
\(=>\orbr{\begin{cases}x=0\\x=5\end{cases}}\) \(\)
\(\)
a) 36x2 - 49=0
(6x)2 - 72 =0
(6x + 7)(6x - 7)=0
suy ra 6x + 7=0 hoặc 6x - 7=0
suy ra x =-7/6 hoặc x=7/6