x^2-16=5x+20
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C
27 tháng 12 2022
\(\left[\left(2x-11\right):3+1\right].5=20\\\left(2x-11\right):3+1=20:5\\ \left(2x-11\right):3+1=4\\ \left(2x-11\right):3=4-1\\ \left(2x-11\right):3=3\\ 2x-11=3.3\\ 2x-11=9\\ 2x=11+9\\ 2x=20\\ x=20:2\\ x=10 \)
27 tháng 12 2022
<=> (2X-11):3+1=20:5
<=> (2X-11):3=4-1
<=> 2X-11=3x3
<=> 2X-11=9
<=> 2X=9+11
<=>X=20:2
<=> X=10
NQ
0
AH
Akai Haruma
Giáo viên
16 tháng 12 2023
1.
PT $\Leftrightarrow 2^{x^2-5x+6}+2^{1-x^2}-2^{7-5x}-1=0$
$\Leftrightarrow (2^{x^2-5x+6}-2^{7-5x})-(1-2^{1-x^2})=0$
$\Leftrightarrow 2^{7-5x}(2^{x^2-1}-1)-(2^{x^2-1}-1)2^{1-x^2}=0$
$\Leftrightarrow (2^{x^2-1}-1)(2^{7-5x}-2^{1-x^2})=0$
$\Rightarrow 2^{x^2-1}-1=0$ hoặc $2^{7-5x}-2^{1-x^2}=0$
Nếu $2^{x^2-1}=1\Leftrightarrow x^2-1=0$
$\Leftrightarrow x^2=1\Leftrightarrow x=\pm 1$
$2^{7-5x}-2^{1-x^2}=0$
$\Leftrightarrow 7-5x=1-x^2\Leftrightarrow x^2-5x+6=0$
$\Leftrightarrow (x-2)(x-3)=0\Leftrightarrow x=2; x=3$
AH
Akai Haruma
Giáo viên
16 tháng 12 2023
2. Đặt $\sin ^2x=a$ thì $\cos ^2x=1-a$. PT trở thành:
$16^a+16^{1-a}=10$
$\Leftrightarrow 16^a+\frac{16}{16^a}=10$
$\Leftrightarrow (16^a)^2-10.16^a+16=0$
Đặt $16^a=x$ thì:
$x^2-10x+16=0$
$\Leftrightarrow (x-2)(x-8)=0$
$\Leftrightarrow x=2$ hoặc $x=8$
$\Leftrightarrow 16^a=2$ hoặc $16^a=8$
$\Leftrightarrow 2^{4a}=2$ hoặc $2^{4a}=2^3$
$\Leftrightarroww 4a=1$ hoặc $4a=3$
$\Leftrightarrow a=\frac{1}{4}$ hoặc $a=\frac{3}{4}$
Nếu $a=\frac{1}{4}\Leftrightarrow \sin ^2x=\frac{1}{4}$
$\Leftrightarrow \sin x=\pm \frac{1}{2}$
Nếu $a=\sin ^2x=\frac{3}{4}\Rightarrow \sin x=\pm \frac{\sqrt{3}}{2}$
Đến đây thì đơn giản rồi.
NT
2
PT
9 tháng 9 2018
1) \(2x^4+3x^3-x^2+3x+2=0\)
\(\Rightarrow2x^4+x^3+2x^3+x^2-2x^2-x+4x+2=0\)
\(\Rightarrow x^3\left(2x+1\right)+x^2\left(2x+1\right)-x\left(2x+1\right)+2\left(2x+1\right)=0\)
\(\Rightarrow\left(2x+1\right)\left(x^3+x^2-x+2\right)=0\)
\(\Rightarrow\left(2x+1\right)\left(x^3+2x^2-x^2-2x+x+2\right)=0\)
\(\Rightarrow\left(2x+1\right)\left[x^2\left(x+2\right)-x\left(x+2\right)+\left(x+2\right)\right]=0\)
\(\Rightarrow\left(2x+1\right)\left(x+2\right)\left(x^2-x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x+1=0\\x+2=0\\x^2-x+1=0\end{matrix}\right.\)
Ta có:
\(x^2-x+1\)
\(=x^2-2x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+1\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)
Vì \(\left(x-\dfrac{1}{2}\right)^2\ge0\) với mọi x
\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\) với mọi x
\(\Rightarrow x^2-x+1\) vô nghiệm
\(\Rightarrow\left[{}\begin{matrix}2x+1=0\\x+2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-2\end{matrix}\right.\)
PT
9 tháng 9 2018
3) \(\left(x+2\right)^4+\left(x+4\right)^4=16\)
Đặt x + 3 = a, ta được
\(\left(a-1\right)^4+\left(a+1\right)^4=16\)
\(\Rightarrow\left[\left(a-1\right)^2\right]^2+\left[\left(a+1\right)^2\right]^2=16\)
\(\Rightarrow\left(a^2-2a+1\right)^2+\left(a^2+2a+1\right)^2=16\)
\(\Rightarrow a^4+4a^2+1+2a^2-4a^3-4a+a^4+4a^2+1+2a^2+4a^3+4a=16\)
\(\Rightarrow2a^4+2.4a^2+2+2.2a^2=16\)
\(\Rightarrow2a^4+8a^2+4a^2+2=16\)
\(\Rightarrow2a^4+12a^2+2-16=0\)
\(\Rightarrow2a^4+12a^2-14=0\)
\(\Rightarrow2a^4-2a^2+14a^2-14=0\)
\(\Rightarrow2a^2\left(a^2-1\right)+14\left(a^2-1\right)=0\)
\(\Rightarrow\left(a^2-1\right)\left(2a^2+14\right)=0\)
\(\Rightarrow\left(a-1\right)\left(a+1\right).2\left(a^2+7\right)=0\)
\(\Rightarrow\left(a-1\right)\left(a+1\right)\left(a^2+7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}a-1=0\\a+1=0\\a^2+7=0\end{matrix}\right.\)
Vì \(a^2\ge0\) với mọi a
\(\Rightarrow a^2+7\ge7\) với mọi a
\(\Rightarrow a^2+7\) vô nghiệm
\(\Rightarrow\left[{}\begin{matrix}a-1=0\\a+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+3-1=0\\x+3+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+2=0\\x+4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-2\\x=-4\end{matrix}\right.\)
H
1
24 tháng 9 2020
a) x2 - 2x + 1 = 16 ( như này chứ nhỉ ? )
<=> x2 - 2x + 1 - 16 = 0
<=> x2 - 2x - 15 = 0
<=> x2 + 3x - 5x - 15 = 0
<=> x( x + 3 ) - 5( x + 3 ) = 0
<=> ( x + 3 )( x - 5 ) = 0
<=> \(\orbr{\begin{cases}x+3=0\\x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\x=5\end{cases}}\)
b) ( 5x + 1 )2 - ( 5x - 3 )( 5x + 3 ) = 30
<=> 25x2 + 10x + 1 - ( 25x2 - 9 ) = 30
<=> 25x2 + 10x + 1 - 25x2 + 9 = 30
<=> 10x + 10 = 30
<=> 10x = 20
<=> x = 2
c) ( x - 1 )( x2 + x + 1 ) - x( x + 2 )( x - 2 ) = 5 ( đã sửa đề )
<=> x3 - 1 - x( x2 - 4 ) = 5
<=> x3 - 1 - x3 + 4x = 5
<=> 4x - 1 = 5
<=> 4x = 6
<=> x = 6/4 = 3/2
15 tháng 2 2020
a. (3x - 1).(2x + 7) - (x + 1).(6x - 5) = 16
<=> 6x^2 + 19x - 7 - (6x^2 + x - 5) = 16
<=> 18x - 2 = 16
<=> 18x = 18
<=> x = 1
b. (10x + 9).x - (5x - 1).(2x + 3) = 8
<=> 10x^2 + 9x - (10x^2 + 13x - 3) = 8
<=> -4x + 3 = 8
<=> -4x = 5
<=> x = -5/4
c. (3x - 5).(7 - 5x) + (5x + 2).(3x - 2) - 2 = 0
<=> -15x^2 + 46x - 35 + 15x^2 - 4x - 4 - 2 = 0
<=> 42x - 41 = 0
<=> x = 41/42
19 tháng 8 2021
1: Ta có: \(\dfrac{5x^2-12}{x^2-1}+\dfrac{3}{x-1}=\dfrac{5x}{x+1}\)
\(\Leftrightarrow\dfrac{5x^2-12}{\left(x-1\right)\left(x+1\right)}+\dfrac{3x+3}{\left(x-1\right)\left(x+1\right)}=\dfrac{5x^2-5x}{\left(x+1\right)\left(x-1\right)}\)
Suy ra: \(5x^2+3x-9=5x^2-5x\)
\(\Leftrightarrow8x=9\)
hay \(x=\dfrac{9}{8}\left(tm\right)\)
2: Ta có: \(\dfrac{3}{x-5}-\dfrac{15-3x}{x^2-25}=\dfrac{3}{x+5}\)
\(\Leftrightarrow\dfrac{3x+15}{\left(x-5\right)\left(x+5\right)}+\dfrac{3x-15}{\left(x-5\right)\left(x+5\right)}=\dfrac{3x-15}{\left(x+5\right)\left(x-5\right)}\)
Suy ra: \(6x=3x-15\)
\(\Leftrightarrow3x=-15\)
hay \(x=-5\left(loại\right)\)
AH
Akai Haruma
Giáo viên
19 tháng 8 2021
2. ĐKXĐ: $x\neq \pm 5$
PT \(\Leftrightarrow \frac{3}{x-5}+\frac{3x-15}{x^2-25}=\frac{3}{x+5}\)
\(\Leftrightarrow \frac{3}{x-5}+\frac{3(x-5)}{(x-5)(x+5)}=\frac{3}{x+5}\)
\(\Leftrightarrow \frac{3}{x-5}+\frac{3}{x+5}=\frac{3}{x+5}\Leftrightarrow \frac{3}{x-5}=0\) (vô lý)
Vậy pt vô nghiệm.
\(\Rightarrow x^2-5x-36=0\Rightarrow x\left(x-9\right)+4\left(x-9\right)=0\Rightarrow\left(x-9\right)\left(x+4\right)=0\Rightarrow\left[{}\begin{matrix}x=9\\x=-4\end{matrix}\right.\)