25-{x+[x+(-8)]}=0
a) 25 - y^2 = 8(x+2009)^2 \Leftrightarrow 8(x+2009)^2 + y^2 = 25
Do y^2 \geq 0 \Rightarrow (x+2009)^2 \leq 25/8
\Rightarrow x+2009 =0 hoặc 1
Nếu x+2009 = 1 \Rightarrow 25 - y^2 = 1\Rightarrow y^2 = 26 (không tìm được y)
Nếu x+2009 = \Rightarrow 25 - y^2 = 0\Rightarrow y^2 = 25, y=5
Vậy (x=0;y=5)
Tìm x, biết:
a) 7x(x + 1) - 3(x + 1) =0
b) 3 ( x + 8) - x^2 - 8x = 0
c) x^2 - 10x = -25
d) x^2 - 10x = -25
a) \(7x\left(x+1\right)-3\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(7x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\7x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{3}{7}\end{matrix}\right.\)
b) 3(x + 8) - x2 - 8x = 0
=> 3(x + 8) - (x2 + 8x) = 0
=> 3(x + 8) - x(x + 8) = 0
=> (x + 8)(3 - x) = 0 => \(\left[{}\begin{matrix}x+8=0\\3-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-8\\x=3\end{matrix}\right.\)
c) \(x^2-10x=-25\Rightarrow x^2-10x+25=0\Rightarrow\left(x-5\right)^2=0\Rightarrow x=5\)
d) Giống câu c
b) 3(x + 8) - x2 - 8x = 0
=> 3(x + 8) - (x2 + 8x) = 0
=> 3(x + 8) - x(x + 8) = 0
=> (x + 8)(3 - x) = 0 =>
c)
Bài 2.Tìm x
a,(x-2018) .3=0
b,25(x-8)=0
c,25+(15+x)=75
a) \(\left(x-2018\right)\cdot3=0\)
\(\Leftrightarrow x-2018=0\)
\(\Leftrightarrow x=2018\)
b) \(25\left(x-8\right)=0\)
\(\Leftrightarrow x-8=0\)
\(\Leftrightarrow x=8\)
c) \(25+\left(15+x\right)=75\)
\(\Leftrightarrow15+x=50\)
\(\Leftrightarrow x=35\)
a) x=2018
b)x=8
c)x=35
Bài 2:
a, (x-2018).3=0
<=> x-2018=0
<=>x=0+2018
<=>x=2018
Vậy x=2018
b, 25(x-8)=0
<=> x-8=0
<=> x=0+8
<=>x=8
Vậy x=8
c, 25+(15+x)=75
<=> 25+15+x=75
<=> 40+x=75
<=>x=75-40
<=>x=35
Vậy x=35
b. x( x – 4) - 2x + 8 = 0
c. x^2-25 –( x+5 ) = 0
d.(2x -1)^2- (4x2 – 1) = 0
e. ( 3x – 1)^2 – ( x +5)^2 = 0
f. x^3 – 8 – (x -2)(x -12) =0
b) x(x-4) - 2x+8 = 0
x(x-4) - 2(x-4) = 0
(x-2) (x-4) = 0
TH1: x-2=0 TH2: x-4=0
x=2 x=4
Vậy x\(\in\){2;4}
\(b,\Leftrightarrow\left(x-4\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\\ c,\Leftrightarrow\left(x-5\right)\left(x+5\right)-\left(x+5\right)=0\\ \Leftrightarrow\left(x+5\right)\left(x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-5\end{matrix}\right.\\ d,\Leftrightarrow\left(2x-1\right)^2-\left(2x-1\right)\left(2x+1\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(2x-1-2x-1\right)=0\\ \Leftrightarrow x=\dfrac{1}{2}\\ e,\Leftrightarrow\left(3x-1-x-5\right)\left(3x-1+x+5\right)=0\\ \Leftrightarrow\left(2x-6\right)\left(4x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\\ f,\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)-\left(x-2\right)\left(x-12\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x^2+x+16\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{63}{4}=0\left(vô.n_0\right)\end{matrix}\right.\\ \Leftrightarrow x=2\)
b) x(x-4)-2x+8=0
x(x-4)-2(x-4)=0
(x-4)(x-2)=0
th1: x-4=0
x=4
th2: x-2=0
x=2
Vậy x thuộc tập hợp 4;-2
(x-2)(4x-20)=0
(x-5)(25-5)=0
(x-4)(2x-8)=0
giúp mình ik ạ
\(\left(x-2\right)\left(4x-20\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\4x-20=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\4x=20\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ \left(x-5\right)\left(25-5x?\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\25-5x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\5x=25\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=5\end{matrix}\right.\\ \left(x-4\right)\left(2x-8\right)\\ \Rightarrow\left[{}\begin{matrix}x-4=0\\2x-8=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4\\2x=8\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=4\end{matrix}\right.\)
a,(x-2)(4x-20)=0
=>x-2=0 hoặc 4x-20=0
=>x=2 hoặc x=5
b,(x-5)(25-5)=0
=>x-5=0 ( vì 25-5 ≠0)
=>x=5
c,(x-4)(2x-8)=0
=>x-4=0 hoặc 2x-8=0
=>x=4
\(\left(x-2\right)\left(4x-20\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\4x-20=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\4x=20\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\)
Vậy x={2;5}
________________________
\(\left(x-5\right)\left(25-5\right)=0\\ \Rightarrow\left(x-5\right)20=0\\ \Rightarrow x-5=0\\ \Rightarrow x=5\)
Vậy x=5
_____________________
\(\left(x-4\right)\left(2x-8\right)=0\\ \Rightarrow2x-8=x-4\\ \Rightarrow2x-8-x=-4\\ \Rightarrow x-8=-4\\ \Rightarrow x=-4+8\\ \Rightarrow x=4\)
96-3(x+8)=42
15.5.(x-25)-225=0
250:x+15=25
36:(x-5)=2^2
[3.(70-x)+5]:2=46
\(a,96-3\left(x+8\right)=42\\ \Rightarrow3\left(x+8\right)=54\\ \Rightarrow x+8=18\\ \Rightarrow x=10.\\ b,15.5\left(x-25\right)-225=0\\ \Rightarrow75\left(x-25\right)-225=0\\ \Rightarrow75\left(x-25\right)=225\\ \Rightarrow x-25=3\\ \Rightarrow x=28.\\ c,250:x+15=25\\ \Rightarrow250:x=10\\ \Rightarrow x=25\\ d,36:\left(x-5\right)=2^2\\ \Rightarrow36:\left(x-5\right)=4\\ \Rightarrow x-5=9\\ \Rightarrow x=14.\\ e,\left[3.\left(70-x\right)+5\right]:2=46\\ \Rightarrow3.\left(70-x\right)+5=92\\ \Rightarrow3\left(70-x\right)=87\\ \Rightarrow70-x=29\\ \Rightarrow x=41.\)
Tìm x :
a, (–31) . (x +7)=0 b, (8 – x) . (x + 13) = 0 c,(x2– 25) . (3– x )=0 d, ( x - 3 ) (x2+4) =0 |
\(a,\left(-31\right).\left(x+7\right)=0\\ \Rightarrow x+7=0\\ \Rightarrow x=-7\\ b,\left(8-x\right).\left(x+13\right)=0\\ \Rightarrow\left[{}\begin{matrix}8-x=0\\x+13=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-13\end{matrix}\right.\\ c,\left(x^2-25\right)\left(3-x\right)=0\\ \Rightarrow\left(x-5\right)\left(x+5\right)\left(3-x\right)=0\\\Rightarrow \left[{}\begin{matrix}x-5=0\\x+5=0\\3-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=-5\\x=3\end{matrix}\right.\\ d,\left(x-3\right)\left(x^2+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-3=0\\x^2+4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x^2=-4\left(loại\right)\end{matrix}\right.\\ \Rightarrow x=3\)
a, (–31) . (x +7)=0
<=> x +7 = 0
<=> x = -7
Vậy x \(\in\left\{-7\right\}\)
b, (8 – x) . (x + 13) = 0
<=> \(\left[{}\begin{matrix}8-x=0\\x+13=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=8\\x=-13\end{matrix}\right.\)
Vậy x \(\in\left\{8;-13\right\}\)
c,(x2– 25) . (3– x )=0
<=> (x - 5) (x + 5) (3 - x) = 0
<=> \(\left[{}\begin{matrix}x-5=0\\x+5=0\\3-x=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=5\\x=-5\\x=3\end{matrix}\right.\)
Vậy x \(\in\left\{5;-5;3\right\}\)
d, ( x - 3 ) (x2 + 4) = 0
<=> \(\left[{}\begin{matrix}x-3=0\\x^2+4=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=3\\x^2=-4\end{matrix}\right.\)(vô lý)
Vậy x \(\in\left\{3\right\}\)
(-7)^3*(25+x)>0
(-8)^2(/x/-12)>0
TA CÓ 25-Y^2=8(X-2019)^2
SUY RA 8(X-2019)^2 LỚN HƠN HOẶC BẰNG 25
SUY RA (X-2019)^2 LỚN HƠN HOẶC BẰNG 25 PHẦN 8
MÀ (X-2019)^2 LÀ SỐ CHÍNH PHƯƠNG
SUY RA (X-2019)^2 =0 HOẶC 1
NẾU (X-2019)^2 =0
SUY RA X-2019=0
SUY RA X=2019
SUY RA 25 -Y^2=0
SUY RA Y^2=25
SUY RA Y=5
NẾU (X-2019)^2 =1
SUY RA X-2019=1
SUY RA X=2020
HOẶC X-2019=-1
SUY RA X= 2018
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