Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Lời giải:
a. $x^2-4x-5=0$
$\Leftrightarrow (x+1)(x-5)=0$
$\Leftrightarrow x+1=0$ hoặc $x-5=0$
$\Leftrightarrow x=-1$ hoặc $x=5$
b.
$5x^2-9x-2=0$
$\Leftrightarrow (x-2)(5x+1)=0$
$\Leftrightarrow x-2=0$ hoặc $5x+1=0$
$\Leftrightarrow x=2$ hoặc $x=\frac{-1}{5}$
c.
$(x^2+1)-5(x^2+1)+6=0$
$\Leftrightarrow a^2-5a+6=0$ (đặt $x^2+1=a$)
$\Leftrightarrow (a-2)(a-3)=0$
$\Leftrightarrow a-2=0$ hoặc $a-3=0$
$\Leftrightarrow x^2-1=0$ hoặc $x^2-2=0$
$\Leftrightarrow (x-1)(x+1)=0$ hoặc $(x-\sqrt{2})(x+\sqrt{2})=0$
$\Leftrightarrow x\in\left\{\pm 1; \pm \sqrt{2}\right\}$
d.
$(x^2+6x)-2(x+3)^2-17=0$
$\Leftrightarrow (x^2+6x+9)-2(x+3)^2-26=0$
$\Leftrightarrow (x+3)^2-2(x+3)^2-26=0$
$\Leftrightarrow -(x+3)^2-26=0$
$\Leftrightarrow (x+3)^2=-26<0$ (vô lý)
Do đó không tồn tại $x$ thỏa mãn.
\(\left(4-3x\right)\left(10x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4-3x=0\\10x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}3x=4\\10x=5\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{4}{3}\\x=\frac{1}{2}\end{cases}}}\)
\(\left(7-2x\right)\left(4+8x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}7-2x=0\\4+8x=0\end{cases}\Rightarrow\orbr{\begin{cases}2x=7\\8x=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}}}\)
rồi thực hiện đến hết ...
Brainchild bé ngây thơ qus e , ko thực hiện đến hết như thế đc đâu :>
\(\left(x-3\right)\left(2x-1\right)=\left(2x-1\right)\left(2x+3\right)\)
\(2x^2-7x+3=4x^2+4x-3\)
\(2x^2-7x+3-4x^2-4x+3=0\)
\(-2x^2-11x+6=0\)
\(2x^2+11x-6=0\)
\(2x^2+12x-x-6=0\)
\(2x\left(x+6\right)-\left(x+6\right)=0\)
\(\left(x+6\right)\left(2x-1\right)=0\)
\(x+6=0\Leftrightarrow x=-6\)
\(2x-1=0\Leftrightarrow2x=1\Leftrightarrow x=\frac{1}{2}\)
\(3x-2x^2=0\)
\(x\left(2x-3\right)=0\)
\(x=0\)
\(2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)
Tự lm tiếp nha
a) 2x (x-5) -(x2-10x +25)=0
\(\Leftrightarrow\)2x(x-5)-(x-5)2=0
\(\Leftrightarrow\)(x-5)(2x-x+5)=0
\(\Leftrightarrow\)(x-5)(x+5)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x-5=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
b) x2 - 9 +3x(x+3) = 0
\(\Leftrightarrow\)(x2 - 9) +3x(x+3) =0
\(\Leftrightarrow\)(x-3)(x+3)+3x(x+3)=0
\(\Leftrightarrow\)(x+3)(x-3+3x)=0
\(\Leftrightarrow\)(x+3)(4x-3)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x+3=0\\4x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=-3\\4x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\frac{3}{4}\end{matrix}\right.\)
c) x3 - 16x = 0
\(\Leftrightarrow\)x(x2-16)=0
\(\Leftrightarrow\)x(x-4)(x+4)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
d) (2x+3)(x-2) - (x2 -4x+4) = 0
\(\Leftrightarrow\)(2x+3)(x-2) -(x-2)2=0
\(\Leftrightarrow\)(x-2)(2x+3-x+2)=0
\(\Leftrightarrow\)(x-2)(x+5)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
e) 9x2 -(x2 -2x +1)=0
\(\Leftrightarrow\)(3x)2-(x-1)2=0
\(\Leftrightarrow\)(3x-x+1)(3x+x-1)=0
\(\Leftrightarrow\)(2x+1)(4x-1)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x+1=0\\4x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x=-1\\4x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=\frac{1}{4}\end{matrix}\right.\)
f)x3-4x2 -9x +36 = 0
\(\Leftrightarrow\)(x3-9x)-(4x2-36)=0
\(\Leftrightarrow\)x(x2-9)-4(x2-9)=0
\(\Leftrightarrow\)(x-4)(x2-9)=0
\(\Leftrightarrow\)(x-4)(x-3)(x+3)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-3=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=4\\x=3\\x=-3\end{matrix}\right.\)
g) 3x - 6 = (x-1).(x-2)
\(\Leftrightarrow\)3(x-2)=(x-1)(x-2)
\(\Leftrightarrow\)x-1=3
\(\Leftrightarrow\)x=4
i) (x-2).(x+2) +(2x+1)2 =-5x.(x-3) =5 (?? đề sao vậy ??)
k) x2 -1 = (x-1).(2x+3)
\(\Leftrightarrow\)(x-1)(x+1)=(x-1)(2x+3)
\(\Leftrightarrow\)x+1=2x+3
\(\Leftrightarrow\)x-2x=3-1
\(\Leftrightarrow\)-x=2
\(\Leftrightarrow\)x=-2
l) (2x-1)2 +(x+3).(x-3) -5x(x-2)=6
\(\Leftrightarrow\)4x2-4x+1+x2-9-5x2+10x=6
\(\Leftrightarrow\)6x-8=6
\(\Leftrightarrow\)6x=14
\(\Leftrightarrow\)x=\(\frac{7}{3}\)
Bài 1.
a) ( 7x - 3 )2 - 5x( 9x + 2 ) - 4x2 = 18
<=> 49x2 - 42x + 9 - 45x2 - 10x - 4x2 = 18
<=> -52x + 9 = 18
<=> -52x = 9
<=> x = -9/52
b) ( x - 7 )2 - 9( x + 4 )2 = 0
<=> x2 - 14x + 49 - 9( x2 + 8x + 16 ) = 0
<=> x2 - 14x + 49 - 9x2 - 72x - 144 = 0
<=> -8x2 - 86x - 95 = 0
<=> -8x2 - 10x - 76x - 95 = 0
<=> -8x( x + 5/4 ) - 76( x + 5/4 ) = 0
<=> ( x + 5/4 )( -8x - 76 ) = 0
<=> \(\orbr{\begin{cases}x+\frac{5}{4}=0\\-8x-76=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{5}{4}\\x=-\frac{19}{2}\end{cases}}\)
c) ( 2x + 1 )2 + ( 4x - 1 )( x + 5 ) = 36
<=> 4x2 + 4x + 1 + 4x2 + 19x - 5 = 36
<=> 8x2 + 23x - 4 - 36 = 0
<=> 8x2 + 23x - 40 = 0
=> Vô nghiệm ( lớp 8 chưa học nghiệm vô tỉ nghen ) :))
Bài 2.
a) x2 - 12x + 39 = ( x2 - 12x + 36 ) + 3 = ( x - 6 )2 + 3 ≥ 3 > 0 ∀ x ( đpcm )
b) 17 - 8x + x2 = ( x2 - 8x + 16 ) + 1 = ( x - 4 )2 + 1 ≥ 1 > 0 ∀ x ( đpcm )
c) -x2 + 6x - 11 = -( x2 - 6x + 9 ) - 2 = -( x - 3 )2 - 2 ≤ -2 < 0 ∀ x ( đpcm )
d) -x2 + 18x - 83 = -( x2 - 18x + 81 ) - 2 = -( x - 9 )2 - 2 ≤ -2 < 0 ∀ x ( đpcm )
1, \(x^3+4x^2+4x=0\Leftrightarrow x\left(x^2+4x+4\right)=0\)
\(\Leftrightarrow x\left(x+2\right)^2=0\Leftrightarrow x=-2;x=0\)
2, \(\left(x+3\right)^2-4=0\Leftrightarrow\left(x+3-2\right)\left(x+3+2\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\Leftrightarrow x=-5;x=1\)
3, \(x^4-9x^2=0\Leftrightarrow x^2\left(x^2-9\right)=0\)
\(\Leftrightarrow x^2\left(x-3\right)\left(x+3\right)=0\Leftrightarrow x=0;\pm3\)
4, \(x^2-6x+9=81\Leftrightarrow\left(x-3\right)^2=9^2\)
\(\Leftrightarrow\left(x-3-9\right)\left(x-3+9\right)=0\Leftrightarrow\left(x-12\right)\left(x+6\right)=0\Leftrightarrow x=-6;x=12\)
5, em xem lại đề nhé
à lag tý @@
5, \(x^3+6x^2+9x-4x=0\Leftrightarrow x^3+6x^2+5x=0\)
\(\Leftrightarrow x\left(x^2+6x+5\right)=0\Leftrightarrow x\left(x^2+x+5x+5\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x+5\right)=0\Leftrightarrow x=-5;x=-1;x=0\)
a.
$x^4-25x^3=0$
$\Leftrightarrow x^3(x-25)=0$
\(\Leftrightarrow \left[\begin{matrix} x^3=0\\ x-25=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=0\\ x=25\end{matrix}\right.\)
b.
$(x-5)^2-(3x-2)^2=0$
$\Leftrightarrow (x-5-3x+2)(x-5+3x-2)=0$
$\Leftrightarrow (-2x-3)(4x-7)=0$
\(\Leftrightarrow \left[\begin{matrix}
-2x-3=0\\
4x-7=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix}
x=\frac{-3}{2}\\
x=\frac{7}{4}\end{matrix}\right.\)
c.
$x^3-4x^2-9x+36=0$
$\Leftrightarrow x^2(x-4)-9(x-4)=0$
$\Leftrightarrow (x-4)(x^2-9)=0$
$\Leftrightarrow (x-4)(x-3)(x+3)=0$
\(\Leftrightarrow \left[\begin{matrix} x-4=0\\ x-3=0\\ x+3=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=4\\ x=3\\ x=-3\end{matrix}\right.\)
d. ĐK: $x\neq 0$
$(-x^3+3x^2-4x):(\frac{-1}{2}x)=0$
$\Leftrightarrow x(-x^2+3x-4):(\frac{-1}{2}x)=0$
$\Leftrightarrow -2(-x^2+3x-4)=0$
$\Leftrightarrow x^2-3x+4=0$
$\Leftrightarrow (x-1,5)^2=-1,75< 0$ (vô lý)
Vậy pt vô nghiệm.
bạn đăng tách ra nhé
a, \(\left(2x+1\right)\left(x-4\right)=\left(2x+1\right)^2\)
\(\Leftrightarrow2x^2-7x-4=4x^2+4x+1\Leftrightarrow2x^2+11x+5=0\)
\(\Leftrightarrow\left(x+5\right)\left(2x+1\right)=0\Leftrightarrow x=-5;x=-\frac{1}{2}\)
b, sửa đề : \(\left(x-4\right)\left(x^2+4x+16\right)-\left(x^2-6\right)=2\)
\(\Leftrightarrow x^3-64-x^2+6=2\Leftrightarrow x^3-x^2-60=0\Leftrightarrow x=4,27...\)
c, \(\left(2x-1\right)^2-\left(3x+4\right)^2=0\Leftrightarrow\left(2x-1+3x+4\right)\left(2x-1-3x-4\right)=0\)
\(\Leftrightarrow\left(5x+3\right)\left(-x-5\right)=0\Leftrightarrow x=-\frac{3}{5};x=-5\)
d, \(\left(9x+2\right)\left(x-1\right)-\left(3x-1\right)^2=0\)
\(\Leftrightarrow9x^2-7x-2-9x^2+6x-1=0\Leftrightarrow-x-3=0\Leftrightarrow x=-3\)
e, \(\left(2x+3\right)^2-4\left(x-1\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow4x^2+12x+9-4\left(x-1\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow4x^2+12x+9-4\left(x^3-x-x^2+1\right)=0\)
\(\Leftrightarrow4x^2+12x+9-4x^3+4x+4x^2-4=0\)
\(\Leftrightarrow-4x^3+8x^2+16x+5=0\Leftrightarrow x=-0,9...;x=-0,41...;x=3,31...\)
f, \(15x\left(x+4-6x-24\right)=0\Leftrightarrow15\left(-5x-20\right)=0\)
\(\Leftrightarrow-75x-300=0\Leftrightarrow x=-4\)
g, \(\left(4x-10\right)\left(2-3x\right)-30^2=0\)
\(\Leftrightarrow8x-12x^2-20+30x-900=0\Leftrightarrow-12x^2+38x-920=0\)
vô nghiệm