\(\left(x^2-1\right)\left(x+1\right)\left(x+3\right)\)

\(=\left(x-1\right)\left(x+1\right)\left(x+1\right)\left(x+3\right)\)

\(=\left(x-1\right)\left(x+1\right)^2\left(x+3\right)\)

\(\left(3k+1\right)\left(4k+2\right)\left(5k+3\right)=192\)

\(\Leftrightarrow60k^3+86k^2+40k+6=192\)

\(\Leftrightarrow60k^3+86k^2+40k-186=0\)

\(\Leftrightarrow2\left(30k^2+73k+93\right)\left(k-1\right)=0\)

\(\Leftrightarrow k=1\)

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Ta có:Nếu k<1, ta có:

\(\left(3k+1\right)\left(4k+2\right)\left(5k+3\right)< \left(3.1+1\right)\left(4.1+2\right)\left(5.1+3\right)=192\left(L\right)\)

Nếu k=1,ta có:

\(\left(3k+1\right)\left(4k+2\right)\left(5k+3\right)=\left(3.1+1\right)\left(4.1+2\right)\left(5.1+3\right)=192\)

Nếu k>1,ta có:

\(\left(3k+1\right)\left(4k+2\right)\left(5k+3\right)>\left(3.1+1\right)\left(4.1+2\right)\left(5.1+3\right)=192\left(L\right)\)

                                         Vậy k=1

pt <=> \(\left(12k^2+10k+2\right)\left(5k+3\right)=192\)

<=>   \(60k^3+86k^2+40k-186=0\)

<=>   \(60k^3-60k^2+146k^2-146k+186k-186=0\)

<=>   \(\left(k-1\right)\left(60k^2+146k+186\right)=0\)

<=>   \(\orbr{\begin{cases}k=1\\60k^2+146k+186=0\end{cases}}\)

TA XÉT TH2: 

=>    \(900k^2+2190k+2790=0\)

<=>   \(\left(30k+36,5\right)^2+1457,75=0\)

DO: \(\left(30k+36,5\right)^2\ge0\forall k\)

=>   \(VT\ge1457,75>0\)

=> pt vô nghiệm

VẬY PT CÓ NGHIỆM DUY NHẤT     \(x=1\)

\(\left(3k+1\right)\left(4k+2\right)\left(5k+3\right)=192\)

\(\Leftrightarrow60k^3+36k^2+30k^2+18k+20k^2+12k+10k+6=192\)

\(\Leftrightarrow60k^3+86k^2+40k-186=0\)

\(\Leftrightarrow2\left(30k^2+73k+93\ne0\right)\left(k-1\right)=0\)

\(\Leftrightarrow k=1\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+1\right)\left(x+3\right)=192\)

\(\Leftrightarrow\left(x^2+2x-3\right)\left(x^2+2x+1\right)=192\)

\(\Leftrightarrow\left(x^2+2x\right)^2-2\left(x^2+2x\right)-3-192=0\)

\(\Leftrightarrow\left(x^2+2x\right)^2-2\left(x^2+2x\right)-195=0\)

\(\Leftrightarrow\left(x^2+2x-15\right)\left(x^2+2x+13\right)=0\)

=>(x+5)(x-3)=0

=>x=3 hoặc x=-5

a) 32-6.(8-2³)+18

= 32 - 6 . (8 - 8) +18

=32 - 0 +18

= 50

b) (3.5-9)³.(1+2.3)²+4²

= (15-9 )³. (1+6)² + 4 ²

= 6³ . 7 ² + 4²

= 216 x 49 + 16

= 10584 + 16

=10600

c) 9234:[3.3.(1+8³)]

= 9234 : [ 3.3.9³]

= 9234 : [ 3.3 . 729]

= 9234 :6561

= 1,407407407

d) 76- {2.[2.5²-(31-2.3)]}+3.25

= 76 - { 2. [ 2.5² - ( 31 - 6)]} + 3.25

= 76 - { 2 . [ 2. 5² - 25 ]} + 3 .25

= 76 - { 2 . [2. 25 - 25]} + 3.25

= 76 - { 2.[ 50 - 25 ]} + 3.25

= 76 - { 2. 25} + 3 . 25

= 76 - 50 + 3. 25

= 76 - 50 + 75

= 101

e) [192-(45:15)⁴  ]:3

= [ 192 - 3 ⁴ ] : 3

= [ 192 - 81 ] : 3

= 111 : 3 

= 37

f) [(192-45:15):3]²

= [ ( 192 - 3) :3 ]²

= [ 189 : 3 ]²

= 63² = 3969

a) (x² - 1)(x² + 4x + 3) = 192

=> (x - 1)(x + 1)(x² + x + 3x + 3) - 192 = 0

=> (x - 1)(x + 1)(x + 1)(x + 3) - 192 = 0

=> [(x - 1)(x + 3)](x + 1)² - 192 = 0

=> (x² + 2x - 3)(x² + 2x + 1) - 192 = 0

Đặt x² + 2x - 3 = k

=> k(k + 4) - 192 = 0

=> k² + 4k - 192 = 0

=> k² + 16k - 12k - 192 = 0

=> k(k + 16) - 12(k + 16) = 0

=> (k - 12)(k + 16) = 0

=> (x² + 2x - 3 - 12)(x² + 2x - 3 + 16) = 0

=> (x² + 2x - 15)(x² + 2x + 13) = 0

=> x² + 5x - 3x - 15 = 0 (do x² + 2x + 13 \(\ne\)0)

=> x(x + 5) - 3(x + 5) = 0

=> (x - 3)(x + 5) = 0

=> \(\orbr{\begin{cases}x-3=0\\x+5=0\end{cases}}\) => \(\orbr{\begin{cases}x=3\\x=-5\end{cases}}\)

a) \(\left(x^2-1\right)\left(x^2+4x+3\right)=192\)

\(\Leftrightarrow x^4+4x^3+3x^2-x^2-4x-3=192\)

\(\Leftrightarrow x^4+4x^3+2x^2-4x-3=192\)

\(\Leftrightarrow x^4+4x^3+2x^2-4x-3-192=0\)

\(\Leftrightarrow x^4+4x^3+2x^2-4x-195=0\)

\(\Leftrightarrow\left(x^3+7x^2+23x+65\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left(x^2+2x+13\right)\left(x+5\right)\left(x-3\right)=0\)

mà \(x^2+2x+13\ne0\) nên: 

\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-5\\x=3\end{cases}}\)

\(y\left(y-4\right)=192\Leftrightarrow y^2-4y+4=196\)\(\Leftrightarrow\left(y-2\right)^2=196=14^2\)

\(\orbr{\begin{cases}y-2=14\\y-2=-14\end{cases}\Rightarrow\orbr{\begin{cases}y=16\\y=-12\left(loai\right)\end{cases}}}\)\(\Rightarrow\orbr{\begin{cases}\left(x+1\right)=4\\\left(x+1\right)=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=-5\end{cases}}}\)