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a: Ta có: \(7x+25=144\)
\(\Leftrightarrow7x=119\)
hay x=17
b: Ta có: \(33-12x=9\)
\(\Leftrightarrow12x=24\)
hay x=2
c: Ta có: \(128-3\left(x+4\right)=23\)
\(\Leftrightarrow3\left(x+4\right)=105\)
\(\Leftrightarrow x+4=35\)
hay x=31
d: Ta có: \(71+\left(726-3x\right)\cdot5=2246\)
\(\Leftrightarrow5\left(726-3x\right)=2175\)
\(\Leftrightarrow726-3x=435\)
\(\Leftrightarrow3x=291\)
hay x=97
e: Ta có: \(720:\left[41-\left(2x+5\right)\right]=40\)
\(\Leftrightarrow41-\left(2x+5\right)=18\)
\(\Leftrightarrow2x+5=23\)
\(\Leftrightarrow2x=18\)
hay x=9
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\(a,\Rightarrow\left(4x-1\right)^2=25=5^2=\left(-5\right)^2\\ \Rightarrow\left[{}\begin{matrix}4x-1=5\\4x-1=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-1\end{matrix}\right.\\ b,\Rightarrow2^x\left(1+2^3\right)=144\\ \Rightarrow2^x=144:9=16=2^4\Rightarrow x=4\\ c,\Rightarrow3^{2x+3}=3^{2\left(x+3\right)}\\ \Rightarrow2x+3=2x+6\Rightarrow0x=3\left(vô.lí\right)\\ \Rightarrow x\in\varnothing\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
a)Câu a sai đề
b)x=4
k mik nha
Học tốt
viết lại đề câu a nha
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Tham khảo:Tìm x thuộc N , biết:a) 2x + 2x+3 =144b) (4x -1)2 =25 x 9 - Hoc24
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-29-9(2x-1)\(^2\)= -110
(=) 9(2x-1)2 = (-29) +110
(=) 9(2x-1)2 = 81
(=) (2x-1)2 =81: 9
(=) (2x-1)2 =9
(=) (2x-1)2 = 32 =(-3)2
\(\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}}\)
\(\orbr{\begin{cases}2x=4\\2x=-2\end{cases}}\)
\(\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)
vậy : ........
a,\(-29-9\left(2x-1\right)^2=-110\)
\(=>-29+110=9.\left(2x-1\right)^2\)
\(=>81=9.\left(2x-1\right)^2\)
\(=>\left(2x-1\right)^2=9\)
\(=>\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}=>\orbr{\begin{cases}x=\frac{4}{2}=2\\x=\frac{-2}{2}=-1\end{cases}}}\)
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a: x=3
b: \(2x-1=2\)
hay \(x=\dfrac{3}{2}\)
c: \(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)
\(a.2^x+2^{x+3}=144\)
\(2^x+2^x.2^3=144\)
\(2^x+2^x.8=144\)
\(2^x.\left(1+8\right)=144\)
\(2^x.9=144\)
\(2^x=144:9\)
\(2^x=16\)
\(2^x=2^4\)
\(\Rightarrow x=4\)
Vậy \(x=4\)
\(b.\left(4.x-1\right)^2=25.9\)
\(\left(4.x-1\right)^2=225^2\)
\(4.x-1=225\)
\(4.x=225+1\)
\(4.x=226\)
\(x=226:4\)
\(x=\frac{226}{4}=\frac{113}{2}\)
Vậy \(x=\frac{113}{2}\)
2x + 2x+3 = 144
<=> 2x + 2x.23 = 144
<=> 2x( 1 + 23 ) = 144
<=> 2x.9 = 144
<=> 2x = 16
<=> 2x = 24
<=> x = 4
( 4x - 1 )2 = 25 . 9
<=> ( 4x - 1 )2 = 52 . 32
<=> ( 4x - 1 )2 = ( 5 . 3 )2
<=> ( 4x - 1 )2 = 152 = (-15)2
<=> \(\hept{\begin{cases}4x-1=15\\4x-1=-15\end{cases}}\Leftrightarrow\hept{\begin{cases}x=4\\x=-\frac{7}{2}\end{cases}}\)