(2x- 5)2 = 49
Rút Gọn:
a) (2x+5)3-(2x-5)3-(120x2+49)
b) (4-5x)2 - (3+5x)2
a) `(2x+5)^3-(2x-5)^3-(120x^2+49)`
`=(2x+5-2x+5)[(2x+5)^2+(2x+5)(2x-5)+(2x-5)^2]-(120x^2+49)`
`=10(12x^2+25)-(120x^2+49)`
`=120x^2+250-120x^2-49`
`=201`
b) `(4-5x)^2-(3+5x)^2=(4-5x+3+5x)(4-5x-3-5x)=7.(-10x+1)=-70x+7`
Lời giải:
a.
$(2x+5)^3-(2x-5)^3-(120x^2+49)$
$=[(2x+5)-(2x-5)][(2x+5)^2+(2x+5)(2x-5)+(2x-5)^2]-(120x^2+49)$
$=10(4x^2+20x+25+4x^2-25+4x^2-20x+25)-(120x^2+49)$
$=10(12x^2+25)-(120x^2+49)=250-49=201$
b.
$(4-5x)^2-(3+5x)^2=[(4-5x)+(3+5x)][(4-5x)-(3+5x)]$
$=7(1-10x)$
a) Ta có: \(\left(2x+5\right)^3-\left(2x-5\right)^3-\left(120x^2+49\right)\)
\(=8x^3+60x^2+150x+125-8x^3+60x^2-150x+125-120x^2-49\)
\(=201\)
b) Ta có: \(\left(4-5x\right)^2-\left(3+5x\right)^2\)
\(=\left(4-5x-3-5x\right)\left(4-5x+3+5x\right)\)
\(=7\left(-10x+1\right)\)
\(=-70x+7\)
tìm x
( 2x +5)^2 -2 (2x +5)(x-1)+ (x+1)^2=49
\(\left(2x+5\right)^2-2\left(2x+5\right)\left(x-1\right)+\left(x+1\right)^2=49\)
<=>\(\left\{\left(2x-5\right)-\left(x-1\right)\right\}^2=49\)
<=> \(\left(2x-5-x+1\right)^2=49\)
<=> \(\left(x-4\right)^2=49\)
<=> \(\hept{\begin{cases}x-4=7\\x-4=-7\end{cases}}\)
<=> \(\hept{\begin{cases}x=11\\x=-3\end{cases}}\)
học tốt
tìm x
a.(2x-5)^2=49
b.(2x+5)^2-(1-2x)^2=10
c.(9-2x)^3=27
a) \(\left(2x-5\right)^2=49\)
\(\left(2x-5\right)^2=\left(\pm7\right)^2\)
\(=>2x-5=7\) hoặc \(2x-5=-7\)
\(\cdot2x-5=7\) \(\cdot2x-5=-7\)
\(2x=5+7\) \(2x=-7+5\)
\(2x=12\) \(2x=-2\)
\(x=12:2\) \(x=-2:2\)
\(x=6\) \(x=-1\)
Vậy x=6 hoặc x=-1
b/ \(\left(2x+5\right)^2-\left(1-2x\right)^2=10\)
\(4x^2+20x+25-\left(1-4x+4x^2\right)=10\)
\(4x^2+20x+25-1+4x-4x^2=10\)
\(24x+24=10\)
\(24x=10-24\)
\(24x=-14\)
\(x=\frac{-14}{24}\)
\(x=\frac{-7}{12}\)
c/ \(\left(9-2x\right)^3=27\)
\(\left(9-2x\right)^3=3^3\)
\(9-2x=3\)
\(2x=9-3\)
\(2x=6\)
\(x=6:2\)
\(x=3\)
2x – 49 = 5 . 3 mũ 2
2x - 49 = 5 . 32
2x - 49 = 5 . 9
2x - 49 = 45
2x = 45 + 49
2x = 94
--> x = 94 : 2 = 47
Vậy x = 47
|2x-5|-7=(1/492)x(1/49-42)...x(1/49-1/20152)
thank các bạn nha
Tìm x,biết:
a)(2x-3)2-49=0
b)2x.(x-5)-7.(5-x)=0
c)x2-3x-10=0
a) \(\Rightarrow\left(2x-3\right)^2=49\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
b) \(\Rightarrow\left(x-5\right)\left(2x+7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-\dfrac{7}{2}\end{matrix}\right.\)
c) \(\Rightarrow x\left(x-5\right)+2\left(x-5\right)=0\Rightarrow\left(x-5\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
a, ⇒ (2x - 3)2 = 49
⇒ (2x - 3)2 = \(\left(\pm7\right)^2\)
⇒ \(\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
b, ⇒ 2x.(x - 5) + 7.(x - 5) = 0
⇒ (x - 5).(2x + 7) = 0
⇒ \(\left[{}\begin{matrix}x-5=0\\2x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\2x=-7\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-\dfrac{7}{2}\end{matrix}\right.\)
c, ⇒ x2 - 5x + 2x - 10 = 0
⇒ (x2 - 5x) + (2x - 10) = 0
⇒ x.(x - 5) +2.(x - 5) = 0
⇒ (x - 5).(x + 2)=0
\(\Rightarrow\left[{}\begin{matrix}x+2=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)
(2x-5)2 =49
(2x-5)2=49 = 72 = (-7)2
TH1: 2x - 5 = 7
=> 2x = 12
=> x = 6
TH2: 2x-5 = -7
=> 2x = -2
=> x = -1
\(\left(2x-5\right)^2=49\)
\(\left(2x-5\right)^2=7^2\)
\(\Rightarrow2x-5=7\)
\(\Rightarrow2x=7+5\)
\(\Rightarrow2x=12\)
\(\Rightarrow x=12:2\)
\(\Rightarrow x=6\)
****
Tìm các số thực x, biết:
a) (2x-3)2-49=0
b) 2x(x-5)-7(5-x)=0
c) x2-3x-10=0
a: \(\left(2x-3\right)^2-49=0\)
\(\Leftrightarrow\left(2x+4\right)\left(2x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)
a. (2x - 3)2 - 49 = 0
<=> (2x - 3)2 - 72 = 0
<=> (2x - 3 + 7)(2x - 3 - 7) = 0
<=> (2x + 4)(2x - 10) = 0
<=> \(\left[{}\begin{matrix}2x+4=0\\2x-10=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)
b. 2x(x - 5) - 7(5 - x) = 0
<=> 2x(x - 5) + 7(x - 5) = 0
<=> (2x + 7)(x - 5) = 0
<=> \(\left[{}\begin{matrix}2x+7=0\\x-5=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\end{matrix}\right.\)
c. x2 - 3x - 10 = 0
<=> x2 - 5x + 2x - 10 = 0
<=> x(x - 5) + 2(x - 5) = 0
<=> (x + 2)(x - 5) = 0
<=> \(\left[{}\begin{matrix}x+2=0\\x-5=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=-2\\x=5\end{matrix}\right.\)
a, (2x - 3)2 - 49 = 0
(2x - 3)2 - 72 = 0
(2x - 3 + 7)( 2x - 3 - 7) = 0
(2x + 4)( 2x - 10) = 0
=> 2x + 4 = 0 => 2x - 10 = 0
2x = - 4 2x = 10
x = - 2 x = 5
1 . Tìm x biết :
a . 2x - ( x + 15 - 29 ) = 17 - 11
b . 49 - ( x - 12 + 54 ) = 2x + 49
c . ( 2x - 5 ) . 2 = 49
d . 2 . | x - 3 | -1 = 6
e . 3 . | x + 1 | - 2 = 4
f . 4 x + 2 + 4 x = 1088
3.|x+1|-2=4
3.|x+1|=4+2
3.|x+1|=6
|x+1|=6:3
|x+1|=2
Trường hợp 1 x+1=2
x=2-1
x=1
trường hợp 2
x+1=-2
x=(-2)-1
x=-3
==> x thuộc {1; -3}
k mk nha chúc học tốt