Tìm x sao cho: (các bạn giải chi tiết hộ mình nhé)
a. 1 - 2x < 7
b. (x - 1) . (x - 2) >0
c. (x -2)^2 . (x + 1) . (x - 4) < 0
d. [x^2 . (x - 3)] : (x - 9) < 0
e. 5/x < 1
Tìm x sao cho: (các bạn giải chi tiết hộ mình nhé)
a. 1 - 2x < 7
b. (x - 1) . (x - 2) >0
c. (x -2)^2 . (x + 1) . (x - 4) < 0
a, 1 - 2x < 7
=> -2x < 6
=> x < -3
=> x thuộc {-4; -5; -6; ...}
b, \(\left(x-1\right)\left(x-2\right)>0\)
th1 :
\(\hept{\begin{cases}x-1< 0\\x-2< 0\end{cases}\Rightarrow\hept{\begin{cases}x< 1\\x< 2\end{cases}\Rightarrow}x< 1\Rightarrow x\in\left\{0;-1;-2;...\right\}}\)
th2 :
\(\hept{\begin{cases}x-1>0\\x-2>0\end{cases}\Rightarrow\hept{\begin{cases}x>1\\x>2\end{cases}\Rightarrow}x>2\Rightarrow x\in\left\{3;4;5;...\right\}}\)
vậy_
c tương tự b
\(a.1-2x< 7\Leftrightarrow2x< 7+1=8\Leftrightarrow x< 8:2\Leftrightarrow x< 4\)
Vậy x < 4
\(b.\left(x-1\right)\left(x-2\right)>0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1>0;x-2>0\\x-1< 0;x-2< 0\end{cases}}\)
\(TH1\Leftrightarrow\orbr{\begin{cases}x-1>0\\x-2>0\end{cases}\Leftrightarrow\orbr{\begin{cases}x>0+1=1\\x>0+2=2\end{cases}\Rightarrow x>2}}\)
\(TH2\Leftrightarrow\orbr{\begin{cases}x-1< 0\\x-2< 0\end{cases}\Leftrightarrow\orbr{\begin{cases}x< 0+1=1\\x< 0+2=2\end{cases}\Rightarrow}}x< 2\)
Vậy \(x\ne2\)
Sửa lại câu a,
\(1-2x< 7\Leftrightarrow2x< 1-7=-6\Leftrightarrow x< -6:2\Leftrightarrow x< -3.\)
c, \(\left(x-2\right)^2\left(x+1\right)\left(x-4\right)< 0\)
\(TH1\hept{\begin{cases}\left(x-2\right)^2< 0\\\left(x+1\right)>0\\\left(x-4\right)>0\end{cases}}\)không xảy ra
\(TH2\hept{\begin{cases}\left(x-2\right)^2>0\\x+1< 0\\x-4>0\end{cases}\Leftrightarrow\hept{\begin{cases}\text{mọi x}\\x< -1\\x>-4\end{cases}}}\Rightarrow-4< x< -1\)
\(TH3\hept{\begin{cases}\left(x-2\right)^2>0\\x+1>0\\x-4< 0\end{cases}\Leftrightarrow\hept{\begin{cases}\text{mọi x}\\x>-1\\x< -4\end{cases}}}\left(loại\right)\)
Vậy -4 < x < -1 hay x thuộc { -2; -3 }
giải các phương trình sau:
a.(x - 1)(x + 2)= 0
b.(x -2)(x -5)=0
c.(x +3)(x -5)=0
d.(x + 1/2)(4x + 4)=0
e.(x -4)(5x -10)=0
f.(2x -1)(3x +6)=0
g.(2,3x -6,9)(0,1x -2)=0
\(a,\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
\(b,\left(x-2\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
\(c,\left(x+3\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)
\(d,\left(x+\dfrac{1}{2}\right)\left(4x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\4x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\4\left(x+1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-1\end{matrix}\right.\)
\(e,\left(x-4\right)\left(5x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\5x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
\(f,\left(2x-1\right)\left(3x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-2\end{matrix}\right.\)
`a,(x-1)(x+2)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
`b,(x -2)(x -5)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
`c,(x +3)(x -5)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)
`d,(x + 1/2)(4x + 4)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\4x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\4x=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-1\end{matrix}\right.\)
`e,(x -4)(5x -10)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\5x-10=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\5x=10\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
`f,(2x -1)(3x +6)=0`
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\3x+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\3x=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-2\end{matrix}\right.\)
`g,(2,3x -6,9)(0,1x -2)=0`
\(\Leftrightarrow\left[{}\begin{matrix}2,3x-6,9=0\\0,1x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2,3x=6,9\\0,1x=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=20\end{matrix}\right.\)
a.(x - 1)(x + 2)= 0
<=> x-1=0 hoặc x+2=0
<=> x=1 hoặc x=-2
b.(x -2)(x -5)=0
<=> x-2=0 hoặc x-5=0
<=> x=2 hoặc x=5
c.(x +3)(x -5)=0
<=> x+3=0 hoặc x-5=0
<=> x=-3 hoặc x=5
d.(x + 1/2)(4x + 4)=0
<=> x+1/2=0 hoặc 4x+4=0
<=> x=-1/2 hoặc x=-1
e.(x -4)(5x -10)=0
<=> x-4=0 hoặc 5x-10=0
<=> x=4 hoặc x=2
f.(2x -1)(3x +6)=0
<=> 2x-1=0 hoặc 3x+6=0
<=> x=1/2 hoặc x=-2
g.(2,3x -6,9)(0,1x -2)=0
<=> 2,3x-6,9=0 hoặc 0,1x-2=0
<=> x=3 hoặc x=20
tìm x biết
a) (2x-3)(2x+3)=0
b) x^2-1=0
c) x^2-9=0
d) 4^2-16=0
e) 25x^2-9=0
a) \(\left(2x-3\right)\left(2x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
b) \(x^2-1=0\Rightarrow\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
c) \(x^2-9=0\Rightarrow\left(x-3\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
d) \(\Rightarrow\left(2x-4\right)\left(2x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
2) \(\Rightarrow\left(5x-3\right)\left(5x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)
Giải các phương trình sau:
a, x2 - 9x +20 = 0
b, x2 - 3x - 18 = 0
c, 2x2 - 9 x + 9 = 0
d, 3x2 - 8x + 4 = 0
e, 3x3 - 6x2 - 9x = 0
f, x(x - 5) - 2 + x = 0
g, x3 + 32 + 6x +8 = 0
h, 2x(x - 2) - 2 + x = 0
i, 5x(1 - x) + x - 1 = 0
k, 4 - 9(x - 1)2 = 0
l, (x - 2)2 - 36(x + 3)2 = 0
\(a)x^2-9x+20=0 \\<=>(x-4)(x-5)=0 \\<=>x=4\ hoặc\ x=5 \\b)x^2-3x-18=0 \\<=>(x+3)(x-6)=0 \\<=>x=-3\ hoặc\ x=6 \\c)2x^2-9x+9=0 \\<=>(x-3)(2x-3)=0 \\<=>x=3\ hoặc\ x=\dfrac{3}{2}\)
d: \(\Leftrightarrow3x^2-6x-2x+4=0\)
=>(x-2)(3x-2)=0
=>x=2 hoặc x=2/3
e: \(\Leftrightarrow3x\left(x^2-2x-3\right)=0\)
=>x(x-3)(x+1)=0
hay \(x\in\left\{0;3;-1\right\}\)
f: \(\Leftrightarrow x^2-5x-2+x=0\)
\(\Leftrightarrow x^2-4x-2=0\)
\(\Leftrightarrow\left(x-2\right)^2=6\)
hay \(x\in\left\{\sqrt{6}+2;-\sqrt{6}+2\right\}\)
Tìm x
a, 3/4x*(x2-9)=0
b, x3-16x=0
c, (x-1)(x+2)-x-2=0
d, 3x3-27x=0
e, x2(x+1)+2x(x+1)=0
f, x(2x-3)-2(3-2x)=0
c: =>(x-1)(x+1)=0
hay \(x\in\left\{1;-1\right\}\)
a,
\(=\dfrac{3}{4x}.\left(x-3\right)\left(x+3\right)\)=0
\(\left\{{}\begin{matrix}\dfrac{3}{4x}=0\\x-3=0\\x+3=0\end{matrix}\right.\)
=>\(x=\left\{3,-3\right\}\)
b,
\(x^3-16x=0\\x\left(x^2-16\right)\\ x\left(x-4\right)\left(x+4\right)\)
\(\left\{{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\)
=>\(x=\left\{-4,0,4\right\}\)
d,
\(3x^3-27x=0\\ 3x\left(x^2-9\right)=0\\ 3x\left(x-3\right)\left(x+3\right)=0\)
\(\left\{{}\begin{matrix}3x=0\\x-3=0\\x+3=0\end{matrix}\right.\)
=>\(x=\left\{-3,0,3\right\}\)
e,
\(x^2+\left(x+1\right)+2x\left(x+1\right)=0\\ x\left(x+1\right)\left(x+2\right)=0\)
\(\left\{{}\begin{matrix}x=0\\x+1=0\\x+2=0\end{matrix}\right.\)
=>\(x=\left\{-2,-1,0\right\}\)
f,
\(x\left(2x-3\right)-2\left(3-2x\right)=0\\ \left(2x-3\right)\left(x+2\right)=0\)
\(\left\{{}\begin{matrix}2x-3=0\\x+2=0\end{matrix}\right.\left\{{}\begin{matrix}x=\dfrac{3}{2}\\x=-2\end{matrix}\right.\)
các bạn giải chi tiết hộ mình nhé
Tìm x, biết
a)x : 9/7 = -3,5
b)0,4 x X -1/5 x =3/4
a) \(x=-\dfrac{3}{5}\times\dfrac{9}{7}=-\dfrac{27}{35}\)
b) \(x\left(0,4-\dfrac{1}{5}\right)=\dfrac{3}{4}\)
\(x=\dfrac{3}{4}:\dfrac{1}{5}=\dfrac{15}{4}\)
a, \(x=-3,5.\dfrac{9}{7}=-\dfrac{9}{2}\)
b, \(\dfrac{2}{5}x-\dfrac{1}{5}x=\dfrac{3}{4}\Leftrightarrow\dfrac{1}{5}x=\dfrac{3}{4}\Leftrightarrow x=\dfrac{3}{4}:\dfrac{1}{5}=\dfrac{15}{4}\)
x=\(\dfrac{9}{7}x-3.5\)
x=\(\dfrac{-9}{5}\)
Giải phương trình
a, (x^2-2)(x^2+x+1)=0
b, 16x^2 - 8x + 5=0
c, 2x^3 - x^2 - 8x + 4=0
d, 3x^3+6x^2 - 75x -150 = 0
e, 2x^5-3x^4+6x^3-8x^2+3=0
*vn:vô nghiệm.
a. \(\left(x^2-2\right)\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2=0\\x^2+x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(vn\right)\end{matrix}\right.\)
\(\Leftrightarrow x=\pm\sqrt{2}\)
-Vậy \(S=\left\{\pm\sqrt{2}\right\}\).
b. \(16x^2-8x+5=0\)
\(\Leftrightarrow16x^2-8x+1+4=0\)
\(\Leftrightarrow\left(4x-1\right)^2+4=0\) (vô lí)
-Vậy S=∅.
c. \(2x^3-x^2-8x+4=0\)
\(\Leftrightarrow x^2\left(2x-1\right)-4\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\pm2\end{matrix}\right.\)
-Vậy \(S=\left\{\dfrac{1}{2};\pm2\right\}\).
d. \(3x^3+6x^2-75x-150=0\)
\(\Leftrightarrow3x^2\left(x+2\right)-75\left(x+2\right)=0\)
\(\Leftrightarrow3\left(x+2\right)\left(x^2-25\right)=0\)
\(\Leftrightarrow3\left(x+2\right)\left(x+5\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\pm5\end{matrix}\right.\)
-Vậy \(S=\left\{-2;\pm5\right\}\)
Giải các phương trình sau:
a/ (3x – 2)(4x + 5) = 0
b/ (2,3x – 6,9)(0,1x + 2) = 0
c/ (4x + 2)(x2 + 1) = 0
d/(2x + 7)(x – 5)(5x + 1) = 0
e/ (x – 1)(2x + 7)(x2 + 2) = 0
f/ (3x + 2)(x2 – 1) = (9x2 – 4)(x + 1)
a) \(\left(3x-2\right)\left(4x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\4x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{2}{3};-\dfrac{5}{4}\right\}\)
b) \(\left(2,3x-6,9\right)\left(0,1x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2,3x-6,9=0\\0,1x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-20\end{matrix}\right.\)
c) \(\left(4x+2\right)\left(x^2+1\right)=0\)
Vì \(x^2+1\ge1>0\forall x\)
\(\Rightarrow4x+2=0\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy: \(S=\left\{-\dfrac{1}{2}\right\}\)
d) \(\left(2x+7\right)\left(x-5\right)\left(5x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+7=0\\x-5=0\\5x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\\x=-\dfrac{1}{5}\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{7}{2};5;-\dfrac{1}{5}\right\}\)
e) \(\left(x-1\right)\left(2x+7\right)\left(x^2+2\right)=0\)
Vì \(x^2+2\ge2>0\forall x\)
\(\Rightarrow\left(x-1\right)\left(2x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\)
f) \(\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\)
\(\Leftrightarrow\left(3x+2\right)\left(x-1\right)\left(x+1\right)-\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[\left(3x+2\right)\left(x+1\right)\right].\left(x-1-3x+2\right)=0\)
\(\Leftrightarrow\left(3x^2+5x+2\right)\left(-2x+1\right)=0\)
\(\Leftrightarrow\left(3x^2+3x+2x+2\right)\left(-2x+1\right)=0\)
\(\Leftrightarrow\left[3x\left(x+1\right)+2\left(x+1\right)\right]\left(-2x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x+2\right)\left(-2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x+2=0\\-2x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{2}{3}\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{-1;-\dfrac{2}{3};\dfrac{1}{2}\right\}\)
tìm x bt
a)x(x-5)-4x+20=0
b)x(x+6)-7x-42=0
c)x^3-5x^2-x+5=0
d)4x^2-25-(2x-5)(3x+7)=0
e)x^3+27+(x+3)(x-9)=0
giúp mk vs ah!!1
a) x(x - 5) - 4x + 20 = 0
\(\Leftrightarrow\) x(x - 5) - (4x + 20)
\(\Leftrightarrow\) x(x - 5) - 4(x - 5) = 0
\(\Leftrightarrow\) (x - 5)(x - 4)
Khi x - 5 = 0 hoặc x - 4 = 0
\(\Leftrightarrow\) x = 5 \(\Leftrightarrow\) x = 4
Vậy S = \(\left\{5;4\right\}\)
b) x(x + 6) - 7x - 42 = 0
\(\Leftrightarrow\) x(x + 6) - (7x - 42) = 0
\(\Leftrightarrow\) x(x + 6) - 7(x + 6) = 0
\(\Leftrightarrow\) (x + 6)(x - 7) = 0
Khi x - 6 = 0 hoặc x - 7 = 0
\(\Leftrightarrow\) x = 6 \(\Leftrightarrow\) x = 7
Vậy S = \(\left\{6;7\right\}\)
c) x3 - 5x2 - x + 5 = 0
\(\Leftrightarrow\) (x3 - 5x2) - (x + 5) = 0
\(\Leftrightarrow\) x2 (x - 5) - (x - 5) = 0
\(\Leftrightarrow\) (x - 5)(x2 - 1) = 0
\(\Leftrightarrow\) (x - 5)(x - 1)(x + 1) = 0
Khi x - 5 = 0 hoặc x - 1 = 0 hoặc x + 1 = 0
\(\Leftrightarrow\) x = 5 \(\Leftrightarrow\) x = 1 \(\Leftrightarrow\) x = -1
Vậy S = \(\left\{5;1;-1\right\}\)
d) 4x2 - 25 - (2x - 5)(3x + 7) = 0
\(\Leftrightarrow\) (2x)2 - 52 - (2x - 5)(3x + 7) = 0
\(\Leftrightarrow\) (2x - 5)(2x + 5) - (2x - 5)(3x + 7) = 0
\(\Leftrightarrow\) (2x - 5) \([\left(2x+5\right)-\left(3x+7\right)]\) = 0
\(\Leftrightarrow\) (2x - 5) ( 2x + 5 - 3x + 7) = 0
\(\Leftrightarrow\) (2x - 5)( -x + 12) = 0
Khi 2x - 5 = 0 hoặc -x + 12 = 0
\(\Leftrightarrow\) 2x = 5 \(\Leftrightarrow\) -x = -12
\(\Leftrightarrow\) x = \(\dfrac{5}{2}\) \(\Leftrightarrow\) x = 12
Vậy S = \(\left\{\dfrac{5}{2};12\right\}\)
e) x3 + 27 + (x + 3)(x - 9) = 0
\(\Leftrightarrow\) x3 - 33 + (x + 3)(x - 9) = 0
\(\Leftrightarrow\) (x - 3)(x2 - 3x + 9) + (x + 3)(x - 9) = 0
\(\Leftrightarrow\) (x - 3) \(\left[\left(x^2-3x+9\right)+\left(x-9\right)\right]\) = 0
\(\Leftrightarrow\) (x - 3) ( x2 - 3x + 9 + x - 9) = 0
\(\Leftrightarrow\) (x - 3)(x2 - 2x) = 0
\(\Leftrightarrow\) (x - 3)x(x - 2)
Khi x - 3 = 0 hoặc x = 0 hoặc x - 2 = 0
\(\Leftrightarrow\) x = 3 \(\Leftrightarrow\) x = 2
Vậy S = \(\left\{3;0;2\right\}\)
Chúc bạn học tốt
a) Ta có: \(x\left(x-5\right)-4x+20=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\end{matrix}\right.\)
b) Ta có: \(x\left(x+6\right)-7x-42=0\)
\(\Leftrightarrow x\left(x+6\right)-7\left(x+6\right)=0\)
\(\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)
c) Ta có: \(x^3-5x^2-x+5=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\\x=-1\end{matrix}\right.\)
d) Ta có: \(4x^2-25-\left(2x-5\right)\left(3x+7\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(2x+5-3x-7\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-2\end{matrix}\right.\)