Tìm x biết x^2 -4x+16=0
Tìm x, biết:
a) ( x 2 - 4x + 16)(x + 4) - x(x + l)(x + 2) + 3 x 2 = 0;
b) (8x + 2)(1 - 3x) + (6x - l)(4x -10) = -50.
a) Thực hiện rút gọn VT = -2x – 64
Giải phương trình -2x – 64 = 0 thu được x = -32.
b) Thực hiện rút gọn VT = -62 x +12
Giải phương trình -62x + 12 = -50 thu được x = 1.
Tìm x
2x-7+(x-14)=0
x^2-6x=0
(x-3)(16-4x)=0
(x-3)-(16-4x)=0
(x-3)+(16-4x)=0
Mấy câu này khá giống nhau nhé anh (câu 1 giống câu 4 và 5, cấu 2 giống câu 3) =)))
Câu 1: 2x - 7 + (x - 14) = 0
<=> 3x -21 = 0
<=> 3x = 21 => x = 7
Câu 2:
x2 - 6x = 0 <=> x.(x - 6) = 0 => \(\orbr{\begin{cases}x=0\\x-6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=6\end{cases}}\)
Chúc anh học tốt !!!
Câu 1, 2 có người làm rồi nên mik làm tiếp cho mấy câu tiếp. Cứ áp dụng A.B = 0 => A = 0 hoặc B = 0
3; ( x - 3 )( 16 - 4x ) = 0
=> x - 3 = 0 hoặc 16 - 4x = 0
=> x = 3 hoặc x = 4
Vậy x = 3 hoặc x = 4.
4; ( x - 3 ) - ( 16 - 4x ) = 0
=> x - 3 - 16 + 4x = 0
=> ( x + 4x ) - ( 3 + 16 ) = 0
=> 5x - 19 = 0
=> x = 19/5
Vậy x = 19/5
5; ( x + 3 ) + ( 16 - 4x ) = 0
=> x + 3 + 16 - 4x = 0
=> ( x - 4x ) + ( 16 + 3 ) = 0
=> 3x + 19 = 0
=> x = 19/3
Vậy x = 19/3
1. Tìm x , biết :
a) x^2 + 10x + 16 = 0
b) 4x^2 - 12x - 7 = 0
a) x2 + 10x + 16 = 0
<=> x2 + 2x + 8x + 16 = 0
<=> x( x + 2 ) + 8( x + 2 ) = 0
<=> ( x + 2 )( x + 8 ) = 0
<=> \(\orbr{\begin{cases}x+2=0\\x+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-8\end{cases}}\)
b) 4x2 - 12x - 7 = 0
<=> 4x2 + 2x - 14x - 7 = 0
<=> 2x( 2x + 1 ) - 7( 2x + 1 ) = 0
<=> ( 2x + 1 )( 2x - 7 ) = 0
<=> \(\orbr{\begin{cases}2x+1=0\\2x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{7}{2}\end{cases}}\)
a. \(x^2+10x+16=0\)
\(\Leftrightarrow x^2+8x+2x+16=0\)
\(\Leftrightarrow x\left(x+8\right)+2\left(x+8\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\x+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-8\end{cases}}\)
b. \(4x^2-12x-7=0\)
\(\Leftrightarrow4x^2+2x-14x-7=0\)
\(\Leftrightarrow2x\left(2x+1\right)-7\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x-7\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-7=0\\2x+1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}2x=7\\2x=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}}\)
Bài làm :
\(\text{a) }x^2+10x+16=0\)
\(\Leftrightarrow x^2+8x+2x+16=0\)
\(\Leftrightarrow x\left(x+8\right)+2\left(x+8\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+8\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\x+8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-8\end{cases}}\)
\(\text{b) }4x^2-12x-7=0\)
\(\Leftrightarrow4x^2+2x-14x-7=0\)
\(\Leftrightarrow2x\left(2x+1\right)-7\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x-7\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-7=0\\2x+1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}2x=7\\2x=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=-\frac{1}{2}\end{cases}.}\)
Tìm x biết :
a) (3x³ + x² – 13x + 5) : (x² + 2x – 1) = 10
b) (x⁴ – 2x² – 8) : (x – 2) = 0
c) (x²-4x) : (x²-8x+16) = 0
\(a,\Leftrightarrow\dfrac{3x^3+6x^2-3x-5x^2-10x+5}{x^2+2x-1}=10\\ \Leftrightarrow\dfrac{3x\left(x^2+2x-1\right)-5\left(x^2+2x-1\right)}{x^2+2x-1}=10\\ \Leftrightarrow3x-5=10\Leftrightarrow3x=15\Leftrightarrow x=5\\ b,\Leftrightarrow\left(x^4+2x^2-4x^2-8\right):\left(x-2\right)=0\\ \Leftrightarrow\left[\left(x^2-4\right)\left(x^2+2\right)\right]:\left(x-2\right)=0\\ \Leftrightarrow\left[\left(x-2\right)\left(x+2\right)\left(x^2+2\right)\right]:\left(x-2\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x^2+2\right)=0\Leftrightarrow x=-2\left(x^2+2>0\right)\\ c,\Leftrightarrow\dfrac{x\left(x-4\right)}{\left(x-4\right)^2}=0\Leftrightarrow\dfrac{x}{x-4}=0\Leftrightarrow x=0\)
b: \(\Leftrightarrow x^4-4x^2+2x^2-8=0\)
hay x=-2
tìm x biết
a, (x^2-4x+16)(x+4)-x(x+1)(x+2)+3x^2=0
b, (8x+2)(1-3x)+(6x-1)(4x-10)=-50
Trả lời:
a, ( x2 - 4x + 16 )( x + 4 ) - x ( x + 1 )( x + 2 ) + 3x2 = 0
<=> x3 + 4x2 - 4x2 - 16x + 16x + 64 - x ( x2 + 3x + 2 ) + 3x2 = 0
<=> x3 + 64 - x3 - 3x2 - 2x + 3x2 = 0
<=> 64 - 2x = 0
<=> 2x = 64
<=> x = 32
Vậy x = 32 là nghiệm của pt.
b, ( 8x + 2 )( 1 - 3x ) + ( 6x - 1 )( 4x - 10 ) = - 50
<=> 8x - 24x2 + 2 - 6x + 24x2 - 60x - 4x + 10 = - 50
<=> - 62x + 12 = - 50
<=> - 62x = - 62
<=> x = 1
Vậy x = 1 là nghiệm của pt.
Tìm x, biết :
a) ( x - 4 )( x^2 + 4x + 16 ) - x( x^2 - 6 ) = 2
b) ( 2x - 1 )^2 - ( 3x + 4 )^2 = 0
a) x^3 - 64 - x^3 +6x = 2
(x^3 - x^3) + 6x = 2+64 quy tắc chuyển vế nhé bạn
6x = 66
x = 66:11
x = 6
Tìm x biết:
a:(x-1)^3-x(x-2)^2-(x-2)=0
b:(2x+5)(2x-7)-(4x+3)^2=16
a) (x - 1)3 - x(x - 2)2 - (x - 2) = 0
<=> x3 - 2x2 + x - x2 + 2x - 1 - x3 + 4x2 - 4x - x + 2 = 0
<=> x2 - 2x + 1 = 0
<=> x2 - 2.x.1 + 12 = 0
<=> (x - 1)2 = 0
x - 1 = 0
x = 0 + 1
x = 1
=> x = 1
a)Ta có : \(\left(x-1\right)^3-x\left(x-2\right)^2-\left(x-2\right)=0\)
\(=>\left(x-1\right)^3-\left(x^2-2x\right)\left(x-2\right)-\left(x-2\right)=0\)
\(=>\left(x-1\right)^3-\left(x-2\right)\left(x^2-2x+1\right)=0\)
\(=>\left(x-1\right)^3-\left(x-2\right)\left(x-1\right)^2=0\)
\(=>\left(x-1\right)^2\left(x-1-x+2\right)=0\)
\(=>\left(x-1\right)^2=0=>x-1=0=>x=1\)
Vậy x=1
b)(2x+5)(2x-7)-(4x+3)2=16
\(=>4x^2-4x-35-16x^2-24x-9-16=0\)
\(=>-\left(12x^2+28x+60\right)=0\)
\(=>12\left(x^2+\frac{7}{3}x+\frac{5}{3}\right)=0\)
\(=>x^2+\frac{7}{3}x+\frac{49}{36}+\frac{11}{36}=0=>\left(x+\frac{7}{6}\right)^2+\frac{11}{36}=0\)
Lại có \(\left(x+\frac{7}{6}\right)^2\ge0=>\left(x+\frac{7}{6}\right)^2+\frac{11}{36}\ge\frac{11}{36}>0\)
Vậy ko có giá trị nào của x thỏa mãn đề bài
\(=>x^2+\frac{7}{3}x+\frac{49}{36}+\frac{11}{36}=0=>\left(x+\frac{7}{6}\right)^2+\frac{11}{36}=0\)
tìm x, biết
x^2 -2x=0
(3x-1)^2-16=0
x^2-25x=0
(4x-1)^2-9=0
a) \(x^2-2x=0\)
\(x\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}}\)
b) \(\left(3x-1\right)^2-16=0\)
\(\left(3x-1\right)^2-4^2=0\)
\(\left(3x-1-4\right)\left(3x-1+4\right)=0\)
\(\left(3x-5\right)\left(3x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-5=0\\3x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=-1\end{cases}}}\)
c) \(x^2-25x=0\)
\(x\left(x-25\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-25=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=25\end{cases}}}\)
d) \(\left(4x-1\right)^2-9=0\)
\(\left(4x-1\right)^2-3^2=0\)
\(\left(4x-1-3\right)\left(4x-1+3\right)=0\)
\(\left(4x-4\right)\left(4x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4x-4=0\\4x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{-1}{2}\end{cases}}}\)
a) \(x^2-2x=0\)
\(x.\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}}\)
vậy..
b) \(\left(3x-1\right)^2-16=0\)
\(\left(3x-1\right)^2=16\)
\(\left(3x-1\right)^2=4^2=\left(-4\right)^2\)
\(\Rightarrow\orbr{\begin{cases}3x-1=4\\3x-1=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=-1\end{cases}}}\)
vậy ...
c) \(x^2-25x=0\)
\(x.\left(x-25\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-25=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=25\end{cases}}}\)
vậy ....
d) \(\left(4x-1\right)^2-9=0\)
\(\left(4x-1\right)^2=3^2=\left(-3\right)^2\)
\(\Rightarrow\orbr{\begin{cases}4x-1=3\\4x-1=-3\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}}\)
vậy ...
a)\(x^2-2x=0\)
\(\Rightarrow x\left(x-2\right)=0\)
\(\Rightarrow\hept{\begin{cases}x=0\\x-2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=0\\x=2\end{cases}}\)
b)\(\left(3x-1\right)^2-16=0\)
\(\left(3x-1-4\right)\left(3x-1+4\right)=0\)
\(\Rightarrow\hept{\begin{cases}3x-1-4=0\\3x-1+4=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{5}{3}\\x=-1\end{cases}}\)
c)\(x^2-25x=0\)
\(x\left(x-25\right)=0\)
\(\Rightarrow\hept{\begin{cases}x=0\\x-25=0\end{cases}\Rightarrow}\hept{\begin{cases}x=0\\x=25\end{cases}}\)
d)\(\left(4x-1\right)^2-9=0\)
\(\left(4x-1-3\right)\left(4x-1+3\right)=0\)
\(\Rightarrow\hept{\begin{cases}4x-1-3=0\\4x-1+3=0\end{cases}\Rightarrow\hept{\begin{cases}x=1\\x=-\frac{1}{2}\end{cases}}}\)
Bài 1:Tìm x biết :
a) x3 - 0,25x = 0 c) x2 - 4x=0 d) 2x2 -4x=0
b) x2 - 10x = -25 d) x2 - 8x+16=0 e/ x2 -16 =0
a)\(x\left(x^2-0,25\right)=0\)
TH1:\(x=0\) TH2:\(x^2-0,25=0\)
\(x^2=0,25=>x=0,5\)
Vậy x E \(\hept{0,5;0}\)