giải dùm vs ạ !
( x -8 ) ( x- 7 ) = 0
( 4x - 24 ) . ( x - 2012 ) = 0
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Những câu hỏi liên quan
H9
HT.Phong (9A5)
CTVHS
20 tháng 10 2023
a) \(3\left(x-5\right)+6=36\)
\(\Rightarrow3\cdot\left(x-5\right)=36-6\)
\(\Rightarrow3\cdot\left(x-5\right)=30\)
\(\Rightarrow x-5=10\)
\(\Rightarrow x=10+5\)
\(\Rightarrow x=15\)
b) \(200-8\cdot\left(x+7\right)=24\)
\(\Rightarrow8\cdot\left(x+7\right)=200-24\)
\(\Rightarrow8\cdot\left(x+7\right)=176\)
\(\Rightarrow x+7=\dfrac{176}{8}\)
\(\Rightarrow x+7=22\)
\(\Rightarrow x=22-7\)
\(\Rightarrow x=15\)
c) \(\left(x-890\right)\left(74-x\right)=0\)
+) \(x-890=0\)
\(\Rightarrow x=890\)
+) \(74-x=0\)
\(\Rightarrow x=74\)
d) \(\left(31-x\right)\cdot\left(x-64\right)=0\)
+) \(31-x=0\)
\(\Rightarrow x=31\)
+) \(x-64=0\)
\(\Rightarrow x=64\)
H9
HT.Phong (9A5)
CTVHS
19 tháng 6 2023
\(\left(x+2\right)-2=0\)
\(\Rightarrow x+2-2=0\)
\(\Rightarrow x=0\)
\(\left(x+3\right)+1=7\)
\(\Rightarrow x+3+1=7\)
\(\Rightarrow x+4=7\)
\(\Rightarrow x=3\)
\(\left(3x-4\right)+4=12\)
\(\Rightarrow3x-4+4=12\)
\(\Rightarrow3x=12\)
\(\Rightarrow x=4\)
\(\left(5x+4\right)-1=13\)
\(\Rightarrow5x+4-1=13\)
\(\Rightarrow5x+3=13\)
\(\Rightarrow5x=10\)
\(\Rightarrow x=2\)
\(\left(4x-8\right)-3=5\)
\(\Rightarrow4x-8-3=5\)
\(\Rightarrow4x-11=5\)
\(\Rightarrow4x=16\)
\(\Rightarrow x=4\)
\(8-\left(2x+4\right)=2\)
\(\Rightarrow8-2x-4=2\)
\(\Rightarrow4-2x=2\)
\(\Rightarrow2x=2\)
\(\Rightarrow x=1\)
\(7+\left(5x+2\right)=14\)
\(\Rightarrow7+5x+2=14\)
\(\Rightarrow9+5x=14\)
\(\Rightarrow5x=5\)
\(\Rightarrow x=1\)
\(5-\left(3x-11\right)=1\)
\(\Rightarrow5-3x+11=1\)
\(\Rightarrow16-3x=1\)
\(\Rightarrow3x=15\)
\(\Rightarrow x=5\)
1 tháng 2 2021
cấy pt dạng ni lớp 8 học rồi mà :v
chỉ là thêm công thức nghiệm vào thôi ._.
1. ( x + 2 )( x + 4 )( x + 6 )( x + 8 ) + 16 = 0
<=> [ ( x + 2 )( x + 8 ) ][ ( x + 4 )( x + 6 ) ] + 16 = 0
<=> ( x2 + 10x + 16 )( x2 + 10x + 24 ) + 16 = 0
Đặt t = x2 + 10x + 16
pt <=> t( t + 8 ) + 16 = 0
<=> t2 + 8t + 16 = 0
<=> ( t + 4 )2 = 0
<=> ( x2 + 10x + 16 + 4 )2 = 0
<=> ( x2 + 10x + 20 )2 = 0
=> x2 + 10x + 20 = 0
Δ' = b'2 - ac = 25 - 20 = 5
Δ' > 0 nên phương trình có hai nghiệm phân biệt
\(x_1=\frac{-b'+\sqrt{\text{Δ}'}}{a}=-5+\sqrt{5}\)
\(x_2=\frac{-b'-\sqrt{\text{Δ}'}}{a}=-5-\sqrt{5}\)
Vậy ...
2. ( x + 1 )( x + 2 )( x + 3 )( x + 4 ) - 24 = 0
<=> [ ( x + 1 )( x + 4 ) ][ ( x + 2 )( x + 3 ) ] - 24 = 0
<=> ( x2 + 5x + 4 )( x2 + 5x + 6 ) - 24 = 0
Đặt t = x2 + 5x + 4
pt <=> t( t + 2 ) - 24 = 0
<=> t2 + 2t - 24 = 0
<=> ( t - 4 )( t + 6 ) = 0
<=> ( x2 + 5x + 4 - 4 )( x2 + 5x + 4 + 6 ) = 0
<=> x( x + 5 )( x2 + 5x + 10 ) = 0
Vì x2 + 5x + 10 có Δ = -15 < 0 nên vô nghiệm
=> x = 0 hoặc x = -5
Vậy ...
3. ( x - 1 )( x - 3 )( x - 5 )( x - 7 ) - 20 = 0
<=> [ ( x - 1 )( x - 7 ) ][ ( x - 3 )( x - 5 ) ] - 20 = 0
<=> ( x2 - 8x + 7 )( x2 - 8x + 15 ) - 20 = 0
Đặt t = x2 - 8x + 7
pt <=> t( t + 8 ) - 20 = 0
<=> t2 + 8t - 20 = 0
<=> ( t - 2 )( t + 10 ) = 0
<=> ( x2 - 8x + 7 - 2 )( x2 - 7x + 8 + 10 ) = 0
<=> ( x2 - 8x + 5 )( x2 - 7x + 18 ) = 0
<=> \(\orbr{\begin{cases}x^2-8x+5=0\\x^2-7x+18=0\end{cases}}\)
+) x2 - 8x + 5 = 0
Δ' = b'2 - ac = 16 - 5 = 11
Δ' > 0 nên có hai nghiệm phân biệt
\(x_1=\frac{-b'+\sqrt{\text{Δ}'}}{a}=-4+\sqrt{11}\)
\(x_2=\frac{-b'+\sqrt{\text{Δ}'}}{a}=-4-\sqrt{11}\)
+) x2 - 7x + 18 = 0
Δ = b2 - 4ac = 49 - 72 = -23 < 0 => vô nghiệm
Vậy ...
DT
0
12 tháng 8 2019
\(a,4x\left(x+1\right)=8\left(x+1\right)\)
\(\Leftrightarrow4x^2+4x-8x-8=0\)
\(\Leftrightarrow4x^2-4x-8=0\)
\(\Leftrightarrow4\left(x^2-x-2\right)=0\)
\(\text{⇔}4\left(x^2-2x+x-2\right)=0\)
\(\text{⇔}4\left(x-2\right)\left(x+1\right)=0\)
\(\text{⇔}\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
\(c,2x\left(x-2\right)-\left(2-x\right)^2=0\)
\(\text{⇔}2x\left(x-2\right)-\left(x-2\right)^2=0\)
\(\text{⇔}\left(x-2\right)\left(2x-x+2\right)=0\)
\(\text{⇔}\left(x-2\right)\left(x+2\right)=0\)
\(\text{⇔}\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
\(d,\left(x-3\right)^3+\left(3-x\right)=0\)
\(\text{⇔}\left(x-3\right)^3-\left(x-3\right)=0\)
\(\text{⇔}\left(x-3\right)\left(x^2-6x+9-1\right)=0\)
\(\text{⇔}\left(x-3\right)\left(x^2-6x+8\right)=0\)
\(\text{⇔}\left(x-3\right)\left(x-2\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\\x=4\end{matrix}\right.\)
\(g,5x\left(x-2000\right)-x+2000=0\)
\(\text{⇔}\left(x-2000\right)\left(5x-1\right)=0\)
\(\text{⇔}\left[{}\begin{matrix}x=2000\\x=\frac{1}{5}\end{matrix}\right.\)
\(n,\left(x+1\right)^2-1+x=0\)
\(\text{⇔}x^2+2x+1-1+x=0\)
\(\text{⇔}x^2+3x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
\(k,\left(1-x\right)^2-1+x=0\)
\(\text{⇔}\left(1-x\right)^2-\left(1-x\right)=0\)
\(\text{⇔}\left(1-x\right)\left(1-x-1\right)=0\)
\(\text{⇔}\left(1-x\right).\left(-x\right)=0\)
\(\text{⇔}\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\)
\(m,x+6x^2=0\)
\(\text{⇔}x\left(1+6x\right)=0\)
\(\text{⇔}\left[{}\begin{matrix}x=0\\x=-\frac{1}{6}\end{matrix}\right.\)
\(h,x^2-4x=0\)
\(\text{⇔}x\left(x-4\right)=0\)
\(\text{⇔}\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
a) (x-8).(x-7) = 0
=> x-8 = 0 => x = 8
x-7=0 => x = 7
KL:...
b) (4x-24).(x-2012) = 0
=> 4x-24 = 0 => 4x = 24 => x= 6
x-2012 = 0 => x = 2012
KL:...
có hai giải pháp bn thấy cái nào đúng thì ủng hộ
\(\left(x-8\right)\cdot\left(x-7\right)=0\)
\(x-8=0\)
\(x=8\)
\(4x-24=4\cdot\left(x-6\right)\)
\(4\cdot\left(x-6\right)\cdot\left(x-2012\right)=0\)
\(x-6=0\)
\(x=6\)
đây là giải pháp 1, giải pháp 2 từ từ mik làm.
chúc cậu học tốt