Bài 1:

a) \(4^{x+2}+4^x=68\)

\(\Rightarrow4^x\cdot\left(4^2+1\right)=68\)

\(\Rightarrow4^x\cdot17=68\)

\(\Rightarrow4^x=\dfrac{68}{17}\)

\(\Rightarrow4^x=4\)

\(\Rightarrow4^x=4^1\)

\(\Rightarrow x=1\)

b) \(5\cdot2^{x+4}-3\cdot2^x=308\)

\(\Rightarrow2^x\cdot\left(5\cdot2^4-3\right)=308\)

\(\Rightarrow2^x\cdot\left(5\cdot16-3\right)=308\)

\(\Rightarrow2^x\cdot77=308\)

\(\Rightarrow2^x=\dfrac{308}{77}\)

\(\Rightarrow2^x=4\)

\(\Rightarrow2^x=2^2\)

\(\Rightarrow x=2\)

c) \(4\cdot3^{x+1}+7\cdot3^x=513\)

\(\Rightarrow3^x\cdot\left(4\cdot3+7\right)=513\)

\(\Rightarrow3^x\cdot19=513\)

\(\Rightarrow3^x=\dfrac{513}{19}\)

\(\Rightarrow3^x=27\)

\(\Rightarrow3^x=3^3\)

\(\Rightarrow x=3\)

d) \(5^{x+4}-5^x=3120\)

\(\Rightarrow5^x\cdot\left(5^4-1\right)=3120\)

\(\Rightarrow5^x\cdot\left(625-1\right)=3120\)

\(\Rightarrow5^x\cdot624=3120\)

\(\Rightarrow5^x\cdot\dfrac{3120}{624}\)

\(\Rightarrow5^x=5\)

\(\Rightarrow5^x=5^1\)

\(\Rightarrow x=1\)

f) \(3\cdot4^{2x+1}-16^x=2816\)

\(\Rightarrow3\cdot4^{2x+1}-\left(4^2\right)^x=2816\)

\(\Rightarrow3\cdot4^{2x+1}-4^{2x}=2816\)

\(\Rightarrow4^{2x}\cdot\left(3\cdot4-1\right)=2816\)

\(\Rightarrow4^{2x}\cdot11=2816\)

\(\Rightarrow4^{2x}=\dfrac{2816}{11}\)

\(\Rightarrow4^{2x}=256\)

\(\Rightarrow\left(2^2\right)^{2x}=2^8\)

\(\Rightarrow2^{4x}=2^8\)

\(\Rightarrow4x=8\)

\(\Rightarrow x=2\)

Bài 2:

\(2^x+124=5^y\)

\(\Rightarrow5^y-2^x=124\)

\(\Rightarrow5^y-2^x=125-1\)

\(\Rightarrow\left\{{}\begin{matrix}5^y=125\\2^x=1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}5^y=5^3\\2^x=2^0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}y=3\\x=0\end{matrix}\right.\)

Vậy: .... 

a) ( x - 3 )² - 4 = 0

<=> ( x - 3 )² - 2² = 0

<=> ( x - 3 - 2 )( x - 3 + 2 ) = 0

<=> ( x - 5 )( x - 1 ) = 0

<=> x = 5 hoặc x = 1

b( 2x + 3 )² - ( 2x + 1 )( 2x - 1 ) = 22

<=> 4x² + 12x + 9 - ( 4x² - 1 ) = 22

<=> 4x² + 12x + 9 - 4x² + 1 = 22

<=> 12x + 10 = 22

<=> 12x = 12

<=> x = 1

c) ( 4x + 3 )( 4x - 3 ) - ( 4x - 5 )² = 16

<=> 16x² - 9 - ( 16x² - 40x + 25 ) = 16

<=> 16x² - 9 - 16x² + 40x - 25 = 16

<=> 40x - 34 = 16

<=> 40x = 50

<=> x = 50/40 = 5/4

d) x³ - 9x² + 27x - 27 = -8

<=> ( x - 3 )³ = -8

<=> ( x - 3 )³ = (-2)³

<=> x - 3 = -2

<=> x = 1 

e) ( x + 1 )³ - x²( x + 3 ) = 2

<=> x³ + 3x² + 3x + 1 - x³ - 3x² = 2

<=> 3x + 1 = 2

<=> 3x = 1

<=> x = 1/3

f) ( x - 2 )³ - x( x - 1 )( x + 1 ) + 6x² = 5

<=> x³ - 6x² + 12x - 8 - x( x² - 1 ) + 6x² = 5

<=> x³ + 12x - 8 - x³ + x = 5

<=> 13x - 8 = 5

<=> 13x = 13

<=> x = 1

a) \(\left(x-3\right)^2-4=0\)

=> \(\left(x-3\right)^2-2^2=0\)

=> \(\left(x-3-2\right)\left(x-3+2\right)=0\)

=> \(\left(x-5\right)\left(x-1\right)=0\)

=> \(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)

b) \(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=22\)

=> \(\left(2x+3\right)^2-\left[\left(2x\right)^2-1^2\right]=22\)

=> \(\left(2x+3\right)^2-\left(4x^2-1\right)=22\)

=> \(\left(2x\right)^2+2\cdot2x\cdot3+3^2-4x^2+1=22\)

=> \(4x^2+12x+9-4x^2+1=22\)

=> \(12x+9+1=22\)

=> \(12x+10=22\)

=> 12x = 12

=> x = 1

c) \(\left(4x+3\right)\left(4x-3\right)-\left(4x-5\right)^2=16\)

=> \(\left(4x\right)^2-3^2-\left[\left(4x\right)^2-2\cdot4x\cdot5+5^2\right]=16\)

=> \(16x^2-9-\left(16x^2-40x+25\right)=16\)

=> \(16x^2-9-16x^2+40x-25=16\)

=> \(-9+40x-25=16\)

=> \(40x=16+25-\left(-9\right)=16+25+9=50\)

=> x = 50/40 = 5/4

d) \(x^3-9x^2+27x-27=-8\)

=> \(x^3-3\cdot x^2\cdot3+3\cdot x\cdot3^2-3^3=8\)

=> \(\left(x-3\right)^3=-8\)

=> \(\left(x-3\right)^3=\left(-2\right)^3\)

=> x - 3  = -2 => x = 1

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

=> \(x^3+3x^2+3x+1-x^3-3x^2=2\)

=> \(3x+1=2\)

=> \(3x=1\)=> x = 1/3

f) \(\left(x-2\right)^3-x\left(x-1\right)\left(x+1\right)+6x^2=5\)

=> \(x^3-3\cdot x^2\cdot2+3\cdot x\cdot2^2-2^3-x\left(x^2-1\right)+6x^2=5\)

=> \(x^3-6x^2+12x-8-x^3+x+6x^2=5\)

=> \(\left(12x+x\right)-8=5\)

=> 13x  = 13

=> x = 1

a) (x+3)^2-4=0

=>(x+3)^2 = 4

=>(x+3)^2 = 2^2 = (-2)^2

=>x+3 = 2 hoặc -2

=> x= -1 hoặc -5

a.  64.4^x=4⁵

4³.4^x=4⁵

4^x=4⁵:4³

4^x=4²

x=2

có ai biết hoặc là fan gacha life không làm ơn kết bạn với mình đi team gacha life của mình đang thiếu người

Thêm nữa câu a) Tính: M(x) + N(x)+ P(x)

B) Tính M(x) - N (x) - P(x)

ok rồi giúp mình với nha

bài làm sai hết rồi!

toán cái gì mà toán 😡

ng ta hỏi bài chứ có lm bài đâu mà sai

\(A=\left(5x^5+5x^4\right):5x^2-\left(2x^4-8x^2-6x+12\right):\left(2x-4\right)\)

Phép chia thứ nhất:

\(\left(5x^5+5x^4\right):5x^2=x^3+x^2\)

Phép chia thứ hai:

Vậy A = ( x^3 + x^2 ) - ( x^3 + 2x^2 - 3 ) = -x^2 + 3

Với x = -2 thì: A = -(-2)^2 + 3 = -4 + 3 = -1

B) bạn làm tương tự nhé

Bài 1

a) \(x=x^5\)

\(x^5-x=0\)

\(x\left(x^4-1\right)=0\)

\(x=0\) hoặc \(x^4-1=0\)

* \(x^4-1=0\)

\(x^4=1\)

\(x=1\)

Vậy x = 0; x = 1

b) \(x^4=x^2\)

\(x^4-x^2=0\)

\(x^2\left(x^2-1\right)=0\)

\(x^2=0\) hoặc \(x^2-1=0\)

*) \(x^2=0\)

\(x=0\)

*) \(x^2-1=0\)

\(x^2=1\)

\(x=1\)

Vậy \(x=0\); \(x=1\)

c) \(\left(x-1\right)^3=x-1\)

\(\left(x-1\right)^3-\left(x-1\right)=0\)

\(\left(x-1\right)\left[\left(x-1\right)^2-1\right]=0\)

\(x-1=0\) hoặc \(\left(x-1\right)^2-1=0\)

*) \(x-1=0\)

\(x=1\)

*) \(\left(x-1\right)^2-1=0\)

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

\(x-1=1\) hoặc \(x-1=-1\)

**) \(x-1=1\)

\(x=2\)

**) \(x-1=-1\)

\(x=0\)

Vậy \(x=0\); \(x=1\); \(x=2\)

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

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

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

\(g\left(x\right)=x^4+x^2-x^3+x-5+5x^3-x^2-3x^4\)

\(g\left(x\right)=\left(x^4-3x^4\right)+\left(5x^3-x^3\right)+\left(x^2-x^2\right)+x-5\)

\(g\left(x\right)=-2x^4+4x^3+x-5\)

`@` `\text {Ans}`

`\downarrow`

`a,`

\(f(x) -3x^2 + x - 1 + x^4 - x^3 - x^2 + 3x^4 + 2x^3\)

`= (x^4 +3x^4) + (-x^3 +2x^3) + (-3x^2 - x^2) + x - 1`

`= 4x^4 + x^3 -4x^2 + x -1`

\(g(x) = x^4 + x^2 - x^3 + x - 5 + 5x^3 - x^2 - 3x^4\)

`= (x^4-3x^4) + (-x^3+5x^3) + (x^2 - x^2) + x -5`

`= -2x^4 + 4x^3 +x - 5`