Làm hộ mình bài 12,13 nhé hjhj hơi mờ 1 chút![]()

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Bài 4:
a: \(x^3=1728\)
=>\(x^3=12^3\)
=>x=12
b: \(\left(x-3\right)^4=1296\)
=>\(\left(x-3\right)^4=6^4\)
=>\(\left[\begin{array}{l}x-3=6\\ x-3=-6\end{array}\right.\Rightarrow\left[\begin{array}{l}x=9\left(nhận\right)\\ x=-3\left(loại\right)\end{array}\right.\)
c: \(\left(3x-1\right)^4=625\)
=>\(\left(3x-1\right)^4=5^4\)
mà 3x-1>=-1 với x∈N
nên 3x-1=5
=>3x=6
=>x=2(nhận)
d: \(x^2\cdot x^4=3^4\cdot9\)
=>\(x^6=3^4\cdot3^2=3^6\)
=>x=3(nhận) hoặc x=-3(loại)
e: \(3^{x-1}=27\)
=>\(3^{x-1}=3^3\)
=>x-1=3
=>x=4
f: \(5^{2x+2}=625\)
=>\(5^{2x+2}=5^4\)
=>2x+2=4
=>2x=2
=>x=1
g: \(8\cdot2^{3x-2}=1024\)
=>\(2^{3x-2}=\frac{1024}{8}=128=2^7\)
=>3x-2=7
=>3x=9
=>x=3
h: \(6^{5-2x}=36^3:6^3=6^3\)
=>5-2x=3
=>2x=5-3=2
=>x=1
Bài 6:
a: \(5^{6x+2}=25^{2x+8}\)
=>\(5^{6x+2}=\left(5^2\right)^{2x+8}=5^{4x+16}\)
=>6x+2=4x+16
=>2x=14
=>x=7
b: \(2^{x}+2^{x+3}=144\)
=>\(2^{x}+2^{x}\cdot8=144\)
=>\(9\cdot2^{x}=144\)
=>\(2^{x}=\frac{144}{9}=16=2^4\)
=>x=4
c: \(x^{2020}=x^{2021}\)
=>\(x^{2021}-x^{2020}=0\)
=>\(x^{2020}\left(x-1\right)=0\)
=>\(\left[\begin{array}{l}x^{2020}=0\\ x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=1\end{array}\right.\)
d: \(\left(2x-2\right)^3=\left(2x-2\right)^{12}\)
=>\(\left(2x-2\right)^{12}-\left(2x-2\right)^3=0\)
=>\(\left(2x-2\right)^3\cdot\left\lbrack\left(2x-2\right)^9-1\right\rbrack=0\)
=>\(\left[\begin{array}{l}2x-2=0\\ \left(2x-2\right)^9=1\end{array}\right.\Rightarrow\left[\begin{array}{l}2x-2=0\\ 2x-2=1\end{array}\right.\)
=>\(\left[\begin{array}{l}2x=2\\ 2x=3\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1\left(nhận\right)\\ x=\frac32\left(loại\right)\end{array}\right.\)
e: \(3^{x}+3^{x+1}+3^{x+2}=1053\)
=>\(3^{x}\left(1+3+3^2\right)=1053\)
=>\(3^{x}=\frac{1053}{13}=81=3^4\)
=>x=4
f: \(3^{x+4}\cdot5^{y}=45^{x}\)
=>\(3^{2x}\cdot5^{x}=3^{x+4}\cdot5^{y}\)
=>2x=x+4 và x=y
=>x=4 và y=x
=>x=4 và y=x=4
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Vì đths đi qua (2;-1) và (-2;5) nên ta có
\(\begin{cases}4a+2b+c=-1\\4a-2b+c=5\end{cases}\)
Trừ theo vế : 4b = -6 <=> b = -3/2
1: Đây là các biểu thức thuộc hằng đằng thức bình phương của tổng
a: \(\left(2x+1\right)^2=\left(2x\right)^2+2\cdot2x\cdot1+1^2=4x^2+4x+1\)
b: \(\left(x+2y\right)^2=x^2+2\cdot x\cdot2y+\left(2y\right)^2=x^2+4xy+4y^2\)
c: \(\left(3x+2y\right)^2=\left(3x\right)^2+2\cdot3x\cdot2y+\left(2y\right)^2=9x^2+12xy+4y^2\)
d: \(\left(2+3m\right)^2=2^2+2\cdot2\cdot3m+\left(3m\right)^2=4+12m+9m^2\)
e: \(\left(3m+4n\right)^2=\left(3m\right)^2+2\cdot3m\cdot4n+\left(4n\right)^2=9m^2+24mn+16n^2\)
f: \(\left(\frac12x+\frac13y\right)^2=\left(\frac12x\right)^2+2\cdot\frac12x\cdot\frac13y+\left(\frac13y\right)^2=\frac14x^2+\frac13xy+\frac19y^2\)
g: \(\left(\frac23x+\frac32y\right)^2=\left(\frac23x\right)^2+2\cdot\frac23x\cdot\frac32y+\left(\frac32y\right)^2=\frac49x^2+2xy+\frac94y^2\)
h: \(\left(\frac32m+\frac12n\right)^2=\left(\frac32m\right)^2+2\cdot\frac32m\cdot\frac12n+\left(\frac12n\right)^2=\frac94m^2+\frac32mn+\frac14n^2\)
i: \(\left(\frac15y+\frac14z\right)^2=\left(\frac15y\right)^2+2\cdot\frac15y\cdot\frac14z+\left(\frac14z\right)^2=\frac{1}{25}y^2+\frac{1}{10}yz+\frac{1}{16}z^2\)
k: \(\left(\frac45x+\frac12y\right)^2=\left(\frac45x\right)^2+2\cdot\frac45x\cdot\frac12y+\left(\frac12y\right)^2=\frac{16}{25}x^2+\frac45xy+\frac14y^2\)
2: Đây là các biểu thức thuộc hằng đằng thức bình phương của hiệu
a: \(\left(2x-1\right)^2=\left(2x\right)^2-2\cdot2x\cdot1+1^2=4x^2-4x+1\)
b: \(\left(x-2y\right)^2=x^2-2\cdot x\cdot2y+\left(2y\right)^2=x^2-4xy+4y^2\)
c: \(\left(3x-2y\right)^2=\left(3x\right)^2-2\cdot3x\cdot2y+\left(2y\right)^2=9x^2-12xy+4y^2\)
d: \(\left(2-3m\right)^2=2^2-2\cdot2\cdot3m+\left(3m\right)^2=4-12m+9m^2\)
e: \(\left(3m-4n\right)^2=\left(3m\right)^2-2\cdot3m\cdot4n+\left(4n\right)^2=9m^2-24mn+16n^2\)
f: \(\left(\frac12x-\frac13y\right)^2=\left(\frac12x\right)^2-2\cdot\frac12x\cdot\frac13y+\left(\frac13y\right)^2=\frac14x^2-\frac13xy+\frac19y^2\)
g: \(\left(\frac23x-\frac32y\right)^2=\left(\frac23x\right)^2-2\cdot\frac23x\cdot\frac32y+\left(\frac32y\right)^2=\frac49x^2-2xy+\frac94y^2\)
h: \(\left(\frac32m-\frac12n\right)^2=\left(\frac32m\right)^2-2\cdot\frac32m\cdot\frac12n+\left(\frac12n\right)^2=\frac94m^2-\frac32mn+\frac14n^2\)
i: \(\left(\frac15y-\frac14z\right)^2=\left(\frac15y\right)^2-2\cdot\frac15y\cdot\frac14z+\left(\frac14z\right)^2=\frac{1}{25}y^2-\frac{1}{10}yz+\frac{1}{16}z^2\)
k: \(\left(\frac45x-\frac12y\right)^2=\left(\frac45x\right)^2-2\cdot\frac45x\cdot\frac12y+\left(\frac12y\right)^2=\frac{16}{25}x^2-\frac45xy+\frac14y^2\)
3: Các biểu thức này thuộc hằng đẳng thức hiệu hai bình phương
a: \(m^2-n^2=\left(m-n\right)\left(m+n\right)\)
b: \(4m^2-16n^2=\left(2m\right)^2-\left(4n\right)^2=\left(2m-4n\right)\left(2m+4n\right)\)
c: \(49-16x^2=\left(7-4x\right)\left(7+4x\right)\)
d: \(25-9y^2=5^2-\left(3y\right)^2=\left(5-3y\right)\left(5+3y\right)\)
e: \(81x^2-16y^2=\left(9x\right)^2-\left(4y\right)^2=\left(9x-4y\right)\left(9x+4y\right)\)
f: \(\frac{1}{16}x^2-\frac{1}{25}y^2=\left(\frac14x-\frac15y\right)\left(\frac14x+\frac15y\right)\)
g: \(\frac{1}{81}x^2-\frac{4}{25}y^2=\left(\frac19x-\frac25y\right)\left(\frac19x+\frac25y\right)\)
h: \(\frac{49}{16}m^2-\frac{16}{25}n^2=\left(\frac74m\right)^2-\left(\frac45n\right)^2=\left(\frac74m-\frac45n\right)\left(\frac74m+\frac45n\right)\)
i: \(\frac{81}{16}x^2-\frac{16}{49}y^2=\left(\frac94x\right)^2-\left(\frac47y\right)^2=\left(\frac94x-\frac47y\right)\left(\frac94x+\frac47y\right)\)
k: \(\frac{16}{81}x^2-\frac{1}{36}y^2=\left(\frac49x-\frac16y\right)\left(\frac49x+\frac16y\right)\)










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