2x+2x+3 =36
(2x-1).(4x2+2x+1)-(3+2x).(3+2x).(3+2x)+36.(x-1).(x-1)
tìm x biết (2x+5).(2x-7)-(2x-3)^2=36
`(2x+5)(2x-7)-(2x-3)^2=36`
`<=>4x^2-14x+10x-35-(4x^2-12x+9)=36`
`<=>4x^2-4x-35-4x^2+12x-9=36`
`<=>8x-44=36`
`<=>8x=80`
`<=>x=10`
Vậy `S={10}`
Ta có: \(\left(2x+5\right)\left(2x-7\right)-\left(2x-3\right)^2=36\)
\(\Leftrightarrow4x^2-14x+10x-35-\left(4x^2-12x+9\right)=36\)
\(\Leftrightarrow4x^2-4x-35-4x^2+12x-9=36\)
\(\Leftrightarrow8x-44=36\)
\(\Leftrightarrow8x=80\)
hay x=10
Vậy: S={10}
24x-4(2x-3/4)-4(3+2x/2)=36-3(x-3/2)-3(3-2x/3)
\(24x-4\left(2x-\frac{3}{4}\right)-4\left(3+\frac{2x}{2}\right)=36-3\left(x-\frac{3}{2}\right)-3\left(3-\frac{2x}{3}\right)\)
Đề như này đúng không bạn
(2x+5)(2x-7)-(2x-3)2 =36
\(\left(2x+5\right)\left(2x-7\right)-\left(2x-3\right)^2=36\)
\(\Leftrightarrow\left(4x^2+10x-14x-35\right)-\left(4x^2-12x+9\right)=36\)
\(\Leftrightarrow\left(4x^2-4x-35\right)-\left(4x^2-12x+9\right)=36\)
\(\Leftrightarrow4x^2-4x-35-4x^2+12x-9=36\)
\(\Leftrightarrow8x-44=36\)
\(\Leftrightarrow8x=80\)
\(\Leftrightarrow x=10\)
Vậy \(x=10\)
Tìm x biết
a.(2x-3)^2=36
b.(2x-3)^2=36
a. (2x-3)2 = 36
(2x-3)2 = 62
=> TH1: 2x - 3 = 6
2x = 9
x = 9/2
TH2: 2x - 3 = -6
2x = -6 + 3
2x = -3
x = -3/2
Vậy x \(\in\){ -3/2 ; 9/2)
Câu b tương tự
a.(2x-3)^2=36
\(\Rightarrow\left(2x-3\right)^2=6^2\)
\(\Rightarrow2x-3=6\)
\(\Rightarrow2x=9\)\(\Rightarrow x=9:2=\frac{9}{2}\)
a)\(\left(2x-3\right)^2=36\)
\(\Rightarrow\left(2x-3\right)^2=6^2\)
\(\Rightarrow\hept{\begin{cases}2x-3=6\\2x-3=-6\end{cases}\Leftrightarrow\hept{\begin{cases}x=\frac{9}{2}\\x=-\frac{3}{2}\end{cases}}}\)
b)
2^2x.3^2x=36^24
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Ta có : \(2^{2x}.3^{2x}=36^{24}\)
\(\Rightarrow\left(2.3\right)^{2x}=\left(6^2\right)^{24}\)
\(\Leftrightarrow6^{2x}=6^{2.24}\)
\(\Rightarrow x=24\)
Chúc học tốt nhé
2x - 2/3 + 1/2x = -1
31/36 - (1/3 - x)2 = 5/6
`2x -2/3 +1/2x =-1`
`=> 2x+1/2x =-1+2/3`
`=> (2+1/2) x =-3/3 + 2/3`
`= 5/2 x = -1/3`
`=> x=-1/3 : 5/2`
`=>x= -2/15`
`31/36 - (1/3-x)^2 =5/6`
`=> (1/3-x)^2 = 31/36 - 5/6`
`=> (1/3-x)^2 =1/36`
`=> (1/3-x)^2 = (+-1/6)^2`
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{3}-x=\dfrac{1}{6}\\\dfrac{1}{3}-x=-\dfrac{1}{6}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\\x=\dfrac{1}{2}\end{matrix}\right.\)
2x+3.(2x-4)=36
2x+3 . ( 2x- 4 ) =36
2x + 6x - 12 = 36
x ( 2 + 6 ) - 12 = 36
8x - 12 = 36
8x = 36 + 12
8x = 48
x = 48 : 8
x = 6
Vậy x = 6