A x 2=4+12+36+...+78732
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a: =>25-4x=1
=>4x=24
hay x=6
b: =>2x-4=0
hay x=2
c: =>x-35=115
hay x=150
d: =>x-2=12
hay x=14
e: =>x-36=216
hay x=252
a.219 - 7(x+1) = 100
7(x+1) = 219 - 100
7(x+1) = 119
x + 1 = 119 : 7
x + 1 = 17
x = 17 - 1
x = 16
b. (3x - 6 ) . 3 = 36
3x - 6 = 36 : 3
3x - 6 = 12
3x = 12 + 6
3x = 18
x = 18 : 3
x = 6
c.716 - ( x-143) = 659
x-143 = 716 - 659
x-143 = 57
x = 57 + 143
x = 200
b. 30 - [4(x-2)+15] = 3
4(x-2) + 15 = 30 - 3
4(x-2)+15 = 27
4(x-2) = 27 - 15
4(x-2) = 12
x-2 = 12 : 4
x-2 = 3
x = 2 + 3 = 5
e.[(8x - 12) : 4] .33 = 36
[(8x - 12) : 4] . 27 = 729
(8x - 12) : 4 = 729 : 27 = 27
8x - 12 = 27 . 4 = 108
8x = 108 + 12 = 120
x = 120 : 8 = 15
a) \(\Leftrightarrow7\left(x+1\right)=119\\ \Leftrightarrow x+1=17\\ \Leftrightarrow x=16\)
b) \(\Leftrightarrow9\left(x-2\right)=36\\ \Leftrightarrow x-2=4\\ \Leftrightarrow x=6\)
c) \(\Leftrightarrow x-143=57\\ \Leftrightarrow x=200\)
d) \(\Leftrightarrow4\left(x-2\right)+15=27\\ \Leftrightarrow4\left(x-2\right)=12\\ \Leftrightarrow x-2=3\\ \Leftrightarrow x=5\)
e) \(\Leftrightarrow\left(2x-3\right).4:4=3^3\\ \Leftrightarrow2x-3=27\\ \Leftrightarrow2x=24\\ \Leftrightarrow x=12\)
Ta có : x4 + x3 + 6x2 + 5x + 5
= (x4 + 5x2) + (x3 + 5x) + (x2 + 5)
= x2(x2 + 5) + x(x2 + 5) + (x2 + 5)
= (x2 + 5)(x2 + x + 1)
2 x 31 x 12+ 4 x 6 x 42 + 8 x 3 x 27 = 24 x 31 + 24 x 42 + 24 x 27 = 24 x (27 + 31 + 42) = 24 x 100 = 2400
b) 36 x 28 + 36 x 82 + 64 x 69 + 64 x 41 = 36 x (28 + 82) + 64 x (69 + 41) = 36 x 110 + 64 x 110 = 110 x (64 + 36) = 110 x 100 = 11000
a ) Đặt \(\sqrt{x+1}=a\Rightarrow x+1=a^2\Rightarrow x=a^2-1\)
Ta có : \(x^2+x+12\sqrt{x+1}=36\)
\(\Leftrightarrow x\left(x+1\right)+12a=36\)
\(\Leftrightarrow a^2\left(a^2-1\right)+12a-36=0\)
\(\Leftrightarrow a^4-a^2+12a-36=0\)
\(\Leftrightarrow a^3\left(a-2\right)+2a^2\left(a-2\right)+3a\left(a-2\right)+18\left(a-2\right)=0\)
\(\Leftrightarrow\left(a-2\right)\left(a^3+2a^2+3a+18\right)=0\)
\(\Leftrightarrow\left(a-2\right)\left[a^2\left(a+3\right)-a\left(a+3\right)+6\left(a+3\right)\right]=0\)
\(\Leftrightarrow\left(a-2\right)\left(a+3\right)\left(a^2-a+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=2\\a=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=2\\\sqrt{x+1}=-3\left(VL\right)\end{matrix}\right.\)
\(\Leftrightarrow x+1=4\Leftrightarrow x=3\)
Vậy ...
b ) \(x^4-8x^2+x+12=0\)
\(\Leftrightarrow\left(x^4-8x^2+16\right)+x-4=0\)
\(\Leftrightarrow\left(x^2-4\right)^2+x-4=0\)
Đặt \(4-x^2=a\) , ta có :
\(a^2+x-4=0\) \(\Rightarrow x=4-a^2\)
Ta có : x = \(4-a^2;a=4-x^2\)
\(\Leftrightarrow x-a=x^2-a^2\)
\(\Leftrightarrow\left(x-a\right)\left(1-x-a\right)=0\)
\(\Leftrightarrow\left(x-4+x^2\right)\left(1-x-4+x^2\right)=0\)
\(\Leftrightarrow\left(x^2+x-4\right)\left(x^2+x-3\right)=0\)
\(\Leftrightarrow...\)