(2x – 3)2 = 49
giúp với
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(2x-49=5.2^3\)
\(2x-49=5.8\)
\(2x-49=40\)
\(2x=89\)
\(x=\frac{89}{2}\)
=.= hok tốt!!
\(2x-49=5.2^3\)
\(2x-49=5.8\)
\(2x-49=40\)
\(2x=40+49\)
\(2x=49\)
\(x=49\div2\)
\(x=44,5\)\(;\frac{89}{2}\)
Vậy \(44,5=\frac{89}{2}\)
:33
\(3x+12=3\left(x-7\right)\)
\(\Leftrightarrow3x+3.4=3\left(x-7\right)\)
\(\Leftrightarrow3\left(x+4\right)=3\left(x-7\right)\)
\(\Leftrightarrow x+4=x-7\)( vô lí )
\(\Rightarrow x\in\left\{\varnothing\right\}\)
\(2x^2-1=49\)
\(\Leftrightarrow2x^2=49+1\)
\(\Leftrightarrow2x^2=50\)
\(\Leftrightarrow x^2=50\div2\)
\(\Leftrightarrow x^2=25\)
\(\Leftrightarrow\sqrt{x^2}=\sqrt{25}\)
\(\Leftrightarrow x=5\)
a. ( 2x + 1 )2 = 49
<=> ( 2x + 1 )2 = 72
<=> 2x + 1 = 7
<=> x = 3
b. ( 2x - 1 )4 = 81
<=> ( 2x - 1 )4 = 34
<=> 2x - 1 = 3
<=> x = 2
c. ( x + 1 )3 = 2x3
<=> x + 1 = 2x
<=> x = 1
d. ( 2x + 1 )3 = 3x3
<=> 2x + 1 = 3x
<=> x = 1
( 2x + 1 )2 = 49
<=> ( 2x + 1 )2 = ( ±7 )2
<=> \(\orbr{\begin{cases}2x+1=7\\2x+1=-7\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-4\end{cases}}\)
( 2x - 1 )4 = 81
<=> ( 2x - 1 )4 = ( ±3 )4
<=> \(\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)
( x + 1 )3 = ( 2x )3
<=> x + 1 = 2x
<=> x - 2x = -1
<=> -x = -1
<=> x = 1
( 2x + 1 )3 = ( 3x )3
<=> 2x + 1 = 3x
<=> 2x - 3x = -1
<=> -x = -1
<=> x = 1
\(\Rightarrow2x-49=160\\ \Rightarrow2x=160+49=209\\ \Rightarrow x=\dfrac{209}{2}=104,5\)
2x+12=3(x-7)
2x+12=3x-21
2x-3x=-21-12
x=-33
=> x=33
a,\(2x+12=3.\left(x-7\right)\)
\(=>2x+12=3x-21\)
\(=>2x-3x=-21-12\)
\(=>-x=-33\)
\(=>x=33\)
b,\(2x^2-1=49\)
\(=>x^2=\frac{50}{2}=25\)
\(=>x=\sqrt{25}=5\)
\(4x^2+8x+5=\) \(\left(2x\right)^2+2.x.2.2+4+1\)
\(=\left(2x+2\right)^2+1\)
với \(x=49\)=> \(\left(49+2\right)^2+1=2602\)
\(x^3+3x^2+3x+1\) \(=\left(x+1\right)^3\)
với \(x=99\)=> \(\left(99+1\right)^3=1000000\)
mấy cau kia làm tương tự nha
Mk chỉ phân tích ra thôi,cn đâu bn tự thay số vào nha!
\(a,A=4x^2+8x+5\)
\(=4x^2+8x+4+1\)
\(=\left(2x+2\right)^2+1\)
\(b,B=x^3+3x^2+3x+1\)
\(=\left(x+1\right)^3\)
\(c,C=x^3-9x^2+27x-26\)
\(=\left(x^3-9x^2+27x-27\right)+1\)
\(=\left(x-3\right)^3+1\)
\(d,D=\left(2x-3\right)^2-\left(4x-6\right)\left(2x-5\right)+\left(2x-5\right)^2\)
\(=\left(2x-3\right)^2-2\left(2x-3\right)\left(2x-5\right)+\left(2x-5\right)^2\)
\(=\left(2x-3-2x+5\right)^2\)
\(=4\)
Vì giá trị của bt ko phụ thuộc vào biến nên bt luôn có giá trị là 4
\(\dfrac{-4}{x}=\dfrac{x}{-49}\\ \Rightarrow x^2=\left(-4\right)\left(-49\right)\\ \Rightarrow x^2=196\\ \Rightarrow x=\pm14\)
\(\dfrac{3.6}{x-3}=\dfrac{5}{3}\\ \Rightarrow5\left(x-3\right)=3.3.6\\ \Rightarrow5\left(x-3\right)=54\\ \Rightarrow x-3=\dfrac{54}{5}\\ \Rightarrow x=\dfrac{54}{5}+3\\ \Rightarrow x=\dfrac{69}{15}\)
\(\left(2x+1\right):2=12:3\\ \left(2x+1\right):2=4\\2x+1=2\\ 2x=1\\ x=\dfrac{1}{2} \)
\(\left(2x-14\right):3=12:9\\ \left(2x-14\right):3=\dfrac{4}{3}\\ 2x-14=4\\ 2x=16\\ x=8\)
Bài làm
1. Rút gọn
\(\frac{2^5.7+2^6}{2^5.5^2-2^5.7}=\frac{2^5\left(7+2\right)}{2^5\left(25-7\right)}=\frac{9}{18}=\frac{1}{2}\)
2. So sánh
Ta có: \(\frac{2.7+6.21+9.28}{4.9+8.27+12.36}\)
\(=\frac{14+126+252}{36+216+432}=\frac{392}{684}=\frac{98}{171}\)
Mà ta thấy, mẫu số của phân số 98/171 lớn hơn mẫu số của phân sốc 49/84 ( 171 > 84 )
=> \(\frac{98}{171}< \frac{49}{84}\)
Vậy \(\frac{2.7+6.21+9.28}{4.9+8.27+12.36}< \frac{49}{84}\)
3. Tìm x
a) ( 2x + 14 ) : 22 - 3 = 1
( 2x + 14 ) : 4 - 3 = 1
( 2x + 14 ) : 4 = 1 + 4
( 2x + 14 ) : 4 = 5
2x + 14 = 5 x 4
2x + 14 = 20
2x = 6
x = 3
Vậy x = 3
b) 32x - 5 + 2 = 29
9x - 5 + 2 = 29
9x - 5 = 27
9x = 32
x = 32 : 9 = \(\frac{32}{9}\)
Vậy x = \(\frac{32}{9}\)
\(=>\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\left[{}\begin{matrix}2x=7+3\\2x=-7+3\end{matrix}\right.\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\left[{}\begin{matrix}x=10:2\\x=-4:2\end{matrix}\right.\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
( 2x - 3)2 = 49
( 2x - 3)2 = 72
\(\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\)
\(\left[{}\begin{matrix}2x=7+3\\2x=-7+3\end{matrix}\right.\)
\(\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=10:2\\x=-4:2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)