a, (4/5)2x+7= 625/256
b, 7x+2 +7x+1 + 7x / 57 = 52x + 2x+1 + 52x+3 / 131
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a: \(4x^3+12=120\)
=>\(4x^3=108\)
=>\(x^3=27=3^3\)
=>x=3
b: \(\left(x-4\right)^2=64\)
=>\(\left[{}\begin{matrix}x-4=8\\x-4=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-4\end{matrix}\right.\)
c: (x+1)^3-2=5^2
=>\(\left(x+1\right)^3=25+2=27\)
=>x+1=3
=>x=2
d: 136-(x+5)^2=100
=>(x+5)^2=36
=>\(\left[{}\begin{matrix}x+5=6\\x+5=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-11\end{matrix}\right.\)
e: \(4^x=16\)
=>\(4^x=4^2\)
=>x=2
f: \(7^x\cdot3-147=0\)
=>\(3\cdot7^x=147\)
=>\(7^x=49\)
=>x=2
g: \(2^{x+3}-15=17\)
=>\(2^{x+3}=32\)
=>x+3=5
=>x=2
h: \(5^{2x-4}\cdot4=10^2\)
=>\(5^{2x-4}=\dfrac{100}{4}=25\)
=>2x-4=2
=>2x=6
=>x=3
i: (32-4x)(7-x)=0
=>(4x-32)(x-7)=0
=>4(x-8)*(x-7)=0
=>(x-8)(x-7)=0
=>\(\left[{}\begin{matrix}x-8=0\\x-7=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=8\\x=7\end{matrix}\right.\)
k: (8-x)(10-2x)=0
=>(x-8)(x-5)=0
=>\(\left[{}\begin{matrix}x-8=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=5\end{matrix}\right.\)
m: \(3^x+3^{x+1}=108\)
=>\(3^x+3^x\cdot3=108\)
=>\(4\cdot3^x=108\)
=>\(3^x=27\)
=>x=3
n: \(5^{x+2}+5^{x+1}=750\)
=>\(5^x\cdot25+5^x\cdot5=750\)
=>\(5^x\cdot30=750\)
=>\(5^x=25\)
=>x=2
\(\frac{7^{x+2}+7^{x+1}+7x}{57}=\frac{5^{2x}+5^{2x+1}+5^{2x+3}}{131}\)
\(\Rightarrow\frac{7x\left(7^2+7^1+1\right)}{57}=\frac{5^{2x}\left(1+5^1+5^3\right)}{131}\)
\(\Rightarrow\frac{7x\left(49+7+1\right)}{57}=\frac{5^{2x}\left(1+5+125\right)}{131}\)
\(\Rightarrow\frac{7x.57}{57}=\frac{5^{2x}.131}{131}\)
\(\Rightarrow7x=25x\)
\(\Rightarrow x=0\)
\(\left(4x-3\right)^4=\left(4x-3\right)^2\)
\(\Rightarrow\left(4x-3\right)^4-\left(4x-3\right)^2=0\)
\(\Rightarrow\left(4x-3\right)^2\left[\left(4x-3\right)^2-1\right]=0\)
\(\Leftrightarrow\hept{\begin{cases}\left(4x-3\right)^2=0\\\left(4x-3\right)^2=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}4x-3=0\\4x-3=-1\\4x-3=1\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=\frac{3}{4}\\x=\frac{1}{2}\\x=1\end{cases}}\)
16. 7 x − 16. 5 2 x = 0
⇔ 7 x = 5 2 x ⇔ 7 / 25 x = 7 / 25 0 ⇔ x = 0
a: 7x+58=100
nên 7x=42
hay x=6
c: x-56:x=16
nên x-14=16
hay x=30
c)x - 56 : 4 = 16
x - 56 = 16 : 4
x- 56 = 4
x =4 + 56
x = 60
d)101 + (36 - 4x) = 105
(36- 4x ) = 105 - 101
36 - 4x = 4
4x = 36 - 4
4x = 32
x = 32:4
x = 8
d, \(=>\left(\frac{4}{5}\right)^{2x+7}=\left(\frac{4}{5}\right)^4.\)
=> \(2x+7=4\)
=> 2x= -3
=> x=-3/2 . Vậy x=-3/2
e, => \(\frac{7^x.7^2+7^x.7+7^x}{57}=\frac{5^{2x}+5^{2x}.5+5^{2x}.5^2}{131}.\)
=> \(\frac{7^x\left(7^2+7+1\right)}{57}=\frac{5^{2x}\left(1+5+5^2\right)}{131}\)
= > \(\frac{7^x.57}{57}=\frac{5^{2x}.131}{131}\)
=> \(7^x=5^{2x}\)
Đến đoạn này là mik nghĩ không ra nhé
Cô làm tiếp giúp Linh Đan:
\(7^x=5^{2x}\Rightarrow7^x=25^x\Rightarrow\frac{7^x}{25^x}=1\Rightarrow\left(\frac{7}{25}\right)^x=1\Rightarrow x=0\)
a) \(\left(x-4\right)^2=\left(x-4\right)^4\)
\(\Rightarrow\left(x-4\right)^2-\left(x-4^4\right)=0\)
\(\Rightarrow\left(x-4\right)^2.\left[1-\left(x-4\right)^2\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-4\right)^2=0\\1-\left(x-4\right)^2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-4=0\\\left(x-4\right)^2=1^2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-4=0\\x-4=1\\x-4=-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=5\\x=3\end{matrix}\right.\)
mk ghi kết quả thôi nhé, nếu từ kết quả mak k biết biến đổi thì ib cho mk
\(x^5-7x^4-x^3+43x^2-36=\left(x-6\right)\left(x-3\right)\left(x-1\right)\left(x+1\right)\left(x+2\right)\)
câu thứ 2 bạn ktra lại đề
\(x^4+2x^3-15x^2-18x+64=\left(x-2\right)\left(x^3+4x^2-7x-32\right)\)
\(x^3-x^2-4=\left(x-2\right)\left(x^2+x+2\right)\)
\(x^3-3x^2-4x+12=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
a) \(x^5-7x^4-x^3+43x^2-36\)
\(=x^3\left(x^2-1\right)-7x^2\left(x^2-1\right)+36\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^3-7x^2+36\right)=\left(x-1\right)\left(x+1\right)\left(x^3+2x^2-9x^2-18x+18x+36\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x^9-9x+18\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x-3\right)\left(x-6\right)\)
c) \(x^4+2x^3-15x^2-18x+64\)
\(=x^3\left(x-2\right)+4x^2\left(x-2\right)-7x\left(x-2\right)-32\left(x-2\right)\)
\(=\left(x-2\right)\left(x^3+4x^2-7x-32\right)\)
a) 2x + 23x = 625
x (2 + 23) = 625
x. 25 = 625
x = 625 : 25
x = 25
b) (2^2+4^2) x +2^4 x 5 * x = 100 (dấu* là dấu nhân vì mik không muốn trùng nhau nên viết vậy)
( 4 + 16 ) x + 2^4 x 5 * x = 100
20 * x + 16 x 5 * x =100
x ( 20 + 16 x 5 ) = 100
x ( 20 + 80 ) = 100
x * 100 =100
x=100 : 100
x= 1
c) ( 3x +1)^2 = 6^2+8^2
(3x +1)^2 = 36 +64
(3x +1)^2 =100
(3x +1)^2 = 10^2
3x + 1 =10
3x=10 - 1=9
x = 9 :3
x = 3
( d và e làm tương tự )
k mik nha!
\(1,\\ a,=7x^3-49x^2+21x\\ b,=x^2-x-42\\ c,=x^2-16x+64\\ d,=9x^2+12x+4\\ e,=x^2-16-25+10x-x^2=10x-41\\ 2,\\ a,\Rightarrow2\left(x-7\right)=19\\ \Rightarrow x-7=\dfrac{19}{2}\Rightarrow x=\dfrac{33}{2}\\ b,\Rightarrow4x^2-20x+25-4x^2+3x-2x=50\\ \Rightarrow-19x=25\Rightarrow x=-\dfrac{25}{19}\)
a: \(\Leftrightarrow2x+7=-4\)
=>2x=-11
hay x=-11/2
b: \(\Leftrightarrow\dfrac{7^x\cdot49+7^x\cdot7+7^x}{57}=\dfrac{5^{2x}+5^{2x+1}+5^{2x+3}}{131}\)
\(\Leftrightarrow7^x=5^{2x}\)
=>x=0