Tìm x:
a) x : 11 = 25;
b) x : 11 = 78
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a: Ta có: \(71-\left(x+33\right)=26\)
\(\Leftrightarrow x+33=45\)
hay x=12
b: Ta có: \(\left(x+73\right)-26=76\)
\(\Leftrightarrow x+73=102\)
hay x=29
c: Ta có: \(45-\left(x+9\right)=6\)
\(\Leftrightarrow x+9=39\)
hay x=30
a) 71 - ( 33 + x ) = 26
(33 + x ) = 71 -26
33 + x = 45
x = 45 - 33 = 12
b) ( x + 73 ) - 26 = 76
( x + 73 ) = 102
x = 102 - 73 = 29
c) 45 - ( x + 9 ) = 6
( x + 9 ) = 45 - 6 = 39
x = 39 - 9 = 30
d) 89 - ( 73 - x ) = 20
73 - x = 89 - 20
73 - x = 69
x = 73 - 69 = 4
e) 4(x+41) = 400
x + 41 = 400 : 4 = 100
x = 100 - 41 = 59
f) 11(x-9 ) = 77
x - 9 = 77 : 11 = 7
x = 7 + 9 = 16
g) x + 7 = 25 + 13 = 38
x = 38 - 7 = 31
h) x + 4 = 198 - 120 = 78
x = 78 - 4 = 74
i) x - 9 = 350 : 5 = 70
x = 70 + 9 = 79
j) 2x - 49 = 5 . 9 = 45
2x = 45 + 49 = 94
x = 94 : 2 = 47
k) 25 + 3( x - 8 ) = 106
3(x-8 ) = 106 - 25 = 81
x - 8 = 81 : 3 = 27
x = 27 +8= 35
l) 9( x + 4 ) - 25 = 20
9( x + 4 ) = 20 + 25 = 45
x + 4 =45 : 9 = 5
x = 5 - 4 = 1
m) 200 - ( 2x + 6 ) = 64
2x + 6 = 200 - 64 = 136
2x = 136 - 6 = 130
x = 130 : 2 = 65
a. `4x^2-20x+25=0`
`<=>(2x)^2-2.2x.5 +5^2=0`
`<=>(2x-5)^2=0`
`<=>2x-5=0`
`<=>x=5/2`
b. `(x-5)(x+5)-(x-3)^2=2(x-7)`
`<=>x^2-25-x^2+6x-9=2x-14`
`<=>6x-34=2x-14`
`<=>4x=20`
`<=>x=5`
\(a,4x^2-20x+25=0\Leftrightarrow\left(2x\right)^2-2.2x.5+5^2=0\)
\(\Leftrightarrow\left(2x-5\right)^2=0\Leftrightarrow x=\dfrac{5}{2}\)
b, \(\left(x-5\right)\left(x+5\right)-\left(x-3\right)^2=2\left(x-7\right)\)
\(\Leftrightarrow x^2-25-x^2+6x-9=2x-14\Leftrightarrow4x=20\Leftrightarrow x=5\)
a) \(\text{5x(x-2)+(2-x)=0}\)
\(\Rightarrow5x\left(x-2\right)-\left(x-2\right)=0\\ \Rightarrow\left(x-2\right)\left(5x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\5x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{5}\end{matrix}\right.\)
b) \(\text{x(2x-5)-10x+25=0}\)
\(\Rightarrow x\left(2x-5\right)-5\left(2x-5\right)=0\\ \Rightarrow\left(x-5\right)\left(2x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\2x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=2,5\end{matrix}\right.\)
c) \(\dfrac{25}{16}-4x^2+4x-1=0\)
\(\Rightarrow\dfrac{9}{16}-4x^2+4x=0\)
\(\Rightarrow-4x^2+4x+\dfrac{9}{16}=0\)
\(\Rightarrow-4x^2-\dfrac{1}{2}x+\dfrac{9}{2}x+\dfrac{9}{16}=0\)
\(\Rightarrow\left(-4x^2-\dfrac{1}{2}x\right)+\left(\dfrac{9}{2}x+\dfrac{9}{16}\right)=0\)
\(\Rightarrow-\dfrac{1}{2}x\left(8x+1\right)+\dfrac{9}{16}\left(8x+1\right)=0\)
\(\Rightarrow\left(-\dfrac{1}{2}x+\dfrac{9}{16}\right)\left(8x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}-\dfrac{1}{2}x+\dfrac{9}{16}=0\\8x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{8}\\x=\dfrac{-1}{8}\end{matrix}\right.\)
a) \(1=\left(2x+0,5\right)^{600}\)
\(\Rightarrow1^{600}=\left(2x+0,5\right)^{600}\)
\(\Rightarrow\left[{}\begin{matrix}2x+0,5=1\\2x+0,5=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=0,5\\2x=-1,5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0,25\\x=-0,75\end{matrix}\right.\)
b) \(\left(x-0,125\right)^2=0,25\)
\(\Rightarrow\left(x-0,125\right)^2=0,5^2\)
\(\Rightarrow\left[{}\begin{matrix}x-0,125=0,5\\x-0,125=-0,5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0,625\\x=-0,375\end{matrix}\right.\)
c) \(\left(x-3\right)^{11}=\left(x-3\right)^{41}\)
\(\Rightarrow\left(x-3\right)^{11}-\left(x-3\right)^{41}=0\)
\(\Rightarrow\left(x-3\right)^{11}\left[1-\left(x-3\right)^{30}\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-3\right)^{11}=0\\\left(x-3\right)^{30}=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x-3=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
`@` `\text {Ans}`
`\downarrow`
`a)`
`1 = (2x + 0,5)^600`
`=> (2x+0,5)^600 = (+-1)^600`
`=> \text {TH1: } 2x + 0,5 = 1`
`=> 2x = 1 - 0,5`
`=> 2x = 0,5`
`=> x = 0,5 \div 2`
`=> x = 0,25`
`\text {TH2: } 2x + 0,5 = -1`
`=> 2x = -1 - 0,5`
`=> 2x = -1,5`
`=> x = -1,5 \div 2`
`=> x = -0,75`
Vậy, `x \in {-0,75; 0,25}.`
`b)`
`(x - 0,125)^2 = 0,25`
`=> (x - 0,125)^2 = (+-0,5)^2`
`=> `\(\left[{}\begin{matrix}x-0,125=0,5\\x-0,125=-0,5\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=0,5+0,125\\x=-0,5+0,125\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=0,625\\x=-0,375\end{matrix}\right.\)
Vậy, `x \in {-0,375; 0,625}.`
`c)`
`(x - 3)^11 = (x - 3)^41`
`=> (x - 3)^11 - (x - 3)^41 = 0`
`=> (x - 3)^11 * [ 1 - (x - 3)^30] = 0`
`=>`\(\left[{}\begin{matrix}\left(x-3\right)^{11}=0\\1-\left(x-3\right)^{30}=0\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x-3=0\\\left(x-3\right)^{30}=1\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=3\\x-3=1\end{matrix}\right.\)
`=>`\(\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)
Vậy, `x \in {3; 4}.`
a. 25 + 3(x-8) = 106
3(x-8) = 106 - 25 = 81
x -8 = 81 : 3 = 27
x = 27 +8 = 35
b. 32(x +4) - 52 = 5.52
9(x+4) - 25 = 5.25 = 125
9(x+4) = 125 + 25 = 150(Đề sai)
\(a,25+3\left(x-8\right)=106\\ 3\left(x-8\right)=81\\ x-8=27\\ x=35\)
\(a,\sqrt{x}=3\Leftrightarrow x=9\\ b,\sqrt{x}=6\Leftrightarrow x=36\\ c,\sqrt{x}=8\Leftrightarrow x=64\\ d,\sqrt{x}=12\Leftrightarrow x=144\\ e,2\sqrt{x}=10\Leftrightarrow\sqrt{x}=5\Leftrightarrow x=25\\ f,3\sqrt{x}=21\Leftrightarrow\sqrt{x}=7\Leftrightarrow x=49\\ g,\sqrt{x}=8\Leftrightarrow x=64\\ h,2+\sqrt{x}=11\Leftrightarrow\sqrt{x}=9\Leftrightarrow x=81\)
a. \(\sqrt{x}=3\)
<=> \(\left(\sqrt{\sqrt{x}}\right)^2-\left(\sqrt{3}\right)^2=0\)
<=> \(\left(\sqrt{\sqrt{x}}-\sqrt{3}\right)\left(\sqrt{\sqrt{x}}+\sqrt{3}\right)=0\)
<=> \(\left[{}\begin{matrix}\sqrt{\sqrt{x}}-\sqrt{3}=0\\\sqrt{\sqrt{x}}+\sqrt{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=9\\x=\left(Vnghiêm\right)\end{matrix}\right.\)
Vậy nghiệm của PT là S = \(\left\{9\right\}\)
\(a,\Leftrightarrow x\left(x+9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-9\end{matrix}\right.\\ b,\Leftrightarrow\left(x+4-4\right)\left(x+4+4\right)=0\\ \Leftrightarrow x\left(x+8\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\\ c,\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\\ d,\Leftrightarrow\left(x-5\right)^2=0\Leftrightarrow x=5\)
a) \(\Leftrightarrow x\left(x+9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-9\end{matrix}\right.\)
b) \(\Leftrightarrow x\left(x+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\)
c) \(\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
d) \(\Leftrightarrow\left(x-5\right)^2=0\\ \Leftrightarrow x=5\)
a) x : 11 = 25
x = 25 x 11
x = 275
b) x : 11 = 78
x = 78 x 11
x = 858