3x : 9 = 391
81 : 3x + 1 = 9
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`Q=(3x-1)(9x^2-3x+1)-(1-3x)(1+3x+9x^2)`
`=(3x-1)(9x^2-3x+1)+(3x-1)(9x^2+3x+1)`
`=(3x-1)(9x^2-3x+1+9x^2+3x+1)`
`=(3x-1)(18x^2+2)`
Thay `x=10` vào biểu thức: `Q=(3.10-1)(18 .10^2+2)=52258`
a) Rút gọn Q = 54 x 3 , thay x = 10 vào tính được Q = 54000;
b) Gợi ý x 4 + y 2 = x + 2 y 4 = 8 4 = 2 . Kết quả P = 2.
a: (x^2+9)(9x^2-1)=0
=>9x^2-1=0
=>x^2=1/9
=>x=1/3 hoặc x=-1/3
b: (4x^2-9)(2^(x-1)-1)=0
=>4x^2-9=0 hoặc 2^(x-1)-1=0
=>x^2=9/4 hoặc x-1=0
=>x=1;x=3/2;x=-3/2
c: (3x+2)(9-x^2)=0
=>(3x+2)(3-x)(3+x)=0
=>\(\left[{}\begin{matrix}3x+2=0\\3-x=0\\3+x=0\end{matrix}\right.\Leftrightarrow x\in\left\{-\dfrac{2}{3};3;-3\right\}\)
d: (3x+3)^2(4x-4^2)=0
=>3x+3=0 hoặc 4x-16=0
=>x=4 hoặc x=-1
e: \(2^{\left(x-5\right)\left(x+2\right)}=1\)
=>(x-5)(x+2)=0
=>x-5=0 hoặc x+2=0
=>x=5 hoặc x=-2
\(\frac{3}{x+1}+\frac{2}{x+2}=\frac{5x+4}{x^2+3x+2}.\)ĐKXĐ: \(x\ne-1;-2\)
\(\Leftrightarrow\frac{3\left(x+2\right)}{\left(x+1\right)\left(x+2\right)}+\frac{2\left(x+1\right)}{\left(x+1\right)\left(x+2\right)}=\frac{5x+4}{\left(x+1\right)\left(x+2\right)}\)
\(\Leftrightarrow3x+6+2x+2=5x+4\)
\(\Leftrightarrow3x+2x-5x=-6-2+4\)
\(\Leftrightarrow0x=-4\)
=> PT vô nghiệm
\(2;\frac{2}{3x-1}-\frac{15}{6x^2-x-1}=\frac{3}{2x-1}\)
\(\Leftrightarrow\frac{2\left(2x-1\right)}{\left(2x-1\right)\left(3x-1\right)}-\frac{15}{6x^2+3x-2x-1}=\frac{3\left(3x-1\right)}{\left(2x-1\right)\left(3x-1\right)}\)
\(\Leftrightarrow\frac{4x-2-15}{\left(2x-1\right)\left(3x-1\right)}=\frac{9x-3}{\left(2x-1\right)\left(3x-1\right)}\)
\(\Leftrightarrow4x-2-15=9x-3\)
\(\Leftrightarrow4x-9x=2+15-3\)
\(\Leftrightarrow-5x=14\)
.....
Đặt \(2x^2-3x+1=t\Rightarrow2x^2-3x-9=t-10\)
Phương trình trở thành:
\(t\left(t-10\right)=-9\Leftrightarrow t^2-10t+9=0\Rightarrow\left[{}\begin{matrix}t=1\\t=9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2-3x+1=1\\2x^2-3x+1=9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2-3x=0\\2x^2-3x-8=0\end{matrix}\right.\)
\(\Leftrightarrow...\) (bấm máy)
a . 3 x = 9 ⇔ 3 x = 3 2 ⇔ x = 2
b . 5 x = 5 3 ⇔ x = 3
c . 3 x + 1 = 3 2 ⇔ x + 1 = 2 ⇔ x = 1
a) Ta có: 3 x = 3 2 nên x = 2.
b) Ta có: 5 x = 5 3 nên x = 3.
c) Ta có: 3 x + 1 = 3 2 nên x +1 = 2, do đó x = 1.
a) 3x – 15 = 25 – 5x
=> 3x + 5x = 25 + 15
=> 8x = 40
=> x = 5
b) 3x - 17 = 2x – 7
=> 3x - 2x = -7 + 17
=> x = 10
c) 2x – 17 = – (3x – 18)
=> 2x - 17 = -3x + 18
=> 2x + 3x = 18 + 17
=> 5x = 35
=> x = 7
d) 3x – 14 = 2(x – 9) + 1
=> 3x - 14 = 2x - 18 + 1
=> 3x - 2x = -18 + 1 + 14
=> x = -3
f) (x – 5)2 = 9
\(\Rightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)
a) Ta có: \(3x-15=25-5x\)
\(\Leftrightarrow3x-15-25+5x=0\)
\(\Leftrightarrow8x-40=0\)
\(\Leftrightarrow8x=40\)
hay x=5
Vậy: x=5
b) Ta có: \(3x-17=2x-7\)
\(\Leftrightarrow3x-17-2x+7=0\)
\(\Leftrightarrow x-10=0\)
hay x=10
Vậy: x=10
c) Ta có: \(2x-17=-\left(3x-18\right)\)
\(\Leftrightarrow2x-17=-3x+18\)
\(\Leftrightarrow2x-17+3x-18=0\)
\(\Leftrightarrow5x-35=0\)
\(\Leftrightarrow5x=35\)
hay x=7
Vậy: x=7
d) Ta có: \(3x-14=2\left(x-9\right)+1\)
\(\Leftrightarrow3x-14=2x-18+1\)
\(\Leftrightarrow3x-14-2x+18-1=0\)
\(\Leftrightarrow x+3=0\)
\(\Leftrightarrow x=-3\)
Vậy: x=-3
f) Ta có: \(\left(x-5\right)^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{2;8\right\}\)
Bài làm
a) 3x : 9 = 391
=> 3x : 32 = 391
=> 3x = 391 x 32
=> 3x = 393
=> x = 393 : 3
=> x = 392
Vậy x = 392
b) 81 : 3x + 1 = 9
81 : 3x = 9 - 1
81 : 3x = 8
3x = 81 : 8
3x = 10,125
x = 10,125 : 3
x = 3,375
Vậy x = 3,375
# Học tốt #
a ) \(3x:9=3^{91}\)
\(\Rightarrow3x:3^2=3^{91}\)
\(3x=3^{91}.3^2\)
\(3x=3^{93}\)
\(x=3^{93}:3\)
\(x=3^{92}\)
Vậy \(x=3^{92}\)
b ) \(81:3x+1=9\)
\(27x+1=9\)
\(27.x=9-1\)
\(27.x=8\)
\(x=8:27\)
\(x=\frac{8}{27}\)
Vậy \(x=\frac{8}{27}\)
Chúc bạn học tốt !!!