Tìm x biết 3 x 2 + 8x + 5 = 0
A. x = - 5 3 ; x = - 1
B. x = - 5 3 ; x = 1
C. x = 5 3 ; x = - 1
D. x = 5 3 ; x = 1
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
x^5 -x^3 -x^2 +1=0
x^3(x^2 -1 )-(x^2-1)=0
(x-1)(x^2+x+1)(x-1)(x+1)=0
(x-1)^2(x+1)(x^2+x+1)=0
=> x=1;x=-1
x^3- 5x^2+ 8x- 4= x^3- x^2- 4x^2+ 4x+ 4x- 4
= x^2(x-1)- 4x(x-1)+4(x-1)
= (x-1)(x^2-4x+4)
= (x-1)(x-1)^2
=(x-1)^3
a/ => x3 = 64 => x3 = 43 => x = 4
b/ => 4x2 - 12x + 9 - x2 - 10x - 25 = 0
=> 3x2 - 22x - 16 = 0
=> (x - 8)(3x + 2) = 0
=> x - 8 = 0 => x = 8
hoặc 3x + 2 = 0 => 3x = -2 => x = -2/3
Vậy x = 8 ; x = -2/3
c/ => x3 - x2 - 4x2 + 8x - 4 = 0
=> x3 - 5x2 + 8x - 4 = 0
=> (x - 2)2 (x - 1) = 0
=> (x - 2)2 = 0 => x - 2 = 0 => x = 2
hoặc x - 1 = 0 => x = 1
Vậy x = 2 ; x = 1
a/
\(x^3-4x^2-\left(x-4\right)=0\)
\(\Leftrightarrow x^2\left(x-4\right)-\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\\x=-1\end{matrix}\right.\)
b/
\(x^5-9x=0\)
\(\Leftrightarrow x\left(x^4-9\right)=x\left(x^2-3\right)\left(x^2+3\right)=0\)
\(\Leftrightarrow x\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\)
c/
\(\left(x^3-x^2\right)^2-4x^2+8x-4=0\)
\(\Leftrightarrow x^4\left(x-1\right)^2-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x^4-4\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x^2-2\right)\left(x^2+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x^2-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\pm\sqrt{2}\end{matrix}\right.\)
a) 2x2+3x-5=0
=> 2x2+5x-2x-5=0
=> x(2x+5)-(2x-5)=0
=> (2x-5)(x-1)=0
=> 2x-5=0, x-1=0
=> x=5/2; 1
\(2x^2+3x-5=0< =>2x^2-2+3x-3=0\)
\(< =>2\left(x+1\right)\left(x-1\right)-3\left(x-1\right)=0\)
\(< =>\left(x-1\right)\left(2x-1\right)=0< =>\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\)
a/\(2x^2+3x-5=0\)
\(\Leftrightarrow2x^2-2x+5x-5=0\)
\(\Leftrightarrow2x\left(x-1\right)+5\left(x-1\right)=0\)
\(\Leftrightarrow\left(2x+5\right)\left(x-1\right)=0\)
=> x=1, x=-5/2
b/\(x^2+2x^2-8x+5=0\)
\(\Leftrightarrow3x^2-8x+5=0\)
\(\Leftrightarrow3x^2-3x-5x+5=0\)
\(\Leftrightarrow\left(3x^2-3x\right)-\left(5x-5\right)=0\)
\(\Leftrightarrow3x\left(x-1\right)-5\left(x-1\right)=0\)
\(\Leftrightarrow\left(3x-5\right)\left(x-1\right)=0\)
=> x=1, x=5/3
Lời giải:
a. $(x^2-9)(5x+15)=0$
$\Rightarrow x^2-9=0$ hoặc $5x+15=0$
Nếu $x^2-9=0$
$\Rightarrow x^2=9=3^2=(-3)^2$
$\Rightarrow x=3$ hoặc $-3$
Nếu $5x+15=0$
$\Rightarrow x=-3$
b.
$x^2-8x=0$
$\Rightarrow x(x-8)=0$
$\Rightarrow x=0$ hoặc $x-8=0$
$\Rightarrow x=0$ hoặc $x=8$
c.
$5+12(x-1)^2=53$
$12(x-1)^2=53-5=48$
$(x-1)^2=48:12=4=2^2=(-2)^2$
$\Rightarrow x-1=2$ hoặc $x-2=-2$
$\Rightarrow x=3$ hoặc $x=0$
d.
$(x-5)^2=36=6^2=(-6)^2$
$\Rightarrow x-5=6$ hoặc $x-5=-6$
$\Rightarrow x=11$ hoặc $x=-1$
e.
$(3x-5)^3=64=4^3$
$\Rightarrow 3x-5=4$
$\Rightarrow 3x=9$
$\Rightarrow x=3$
f.
$4^{2x}+2^{4x+3}=144$
$2^{4x}+2^{4x}.8=144$
$2^{4x}(1+8)=144$
$2^{4x}.9=144$
$2^{4x}=144:9=16=2^4$
$\Rightarrow 4x=4\Rightarrow x=1$
a) \(\left|2x+1\right|=\left|1-x\right|\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=1-x\\2x+1=x-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}3x=0\\x=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)
b) \(\left|5x-4\right|=\left|x+2\right|\)
\(\Leftrightarrow\orbr{\begin{cases}5x-4=x+2\\5x-4=-x-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}4x=6\\6x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{1}{3}\end{cases}}\)
c) \(\left|2x-3\right|-\left|3x+2\right|=0\Leftrightarrow\left|2x-3\right|=\left|3x+2\right|\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=3x+2\\2x-3=-3x-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-5\\5x=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=-5\\x=\frac{1}{5}\end{cases}}\)
d) \(\left|2+3\right|=\left|4x-3\right|\Leftrightarrow\left|4x-3\right|=5\)
\(\Rightarrow\orbr{\begin{cases}4x-3=5\\4x-3=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}4x=8\\4x=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=-\frac{1}{2}\end{cases}}\)
e) \(\left|\frac{5}{4}-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\Leftrightarrow\left|\frac{5}{8}x+\frac{3}{5}\right|=\frac{9}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{5}{8}x+\frac{3}{5}=\frac{9}{4}\\\frac{5}{8}x+\frac{3}{5}=-\frac{9}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{5}{8}x=\frac{33}{20}\\\frac{5}{8}x=-\frac{57}{20}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{66}{25}\\x=-\frac{114}{25}\end{cases}}\)
\(\left|2x+1\right|=\left|1-x\right|\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=-x+1\\2x+1=x-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x+x=-1+1\\2x-x=-1-1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x=0\\x=-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)
b. \(\left|5x-4\right|=\left|x+2\right|\)
\(\Leftrightarrow\orbr{\begin{cases}5x-4=x+2\\5x-4=-x-2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}5x-x=4+2\\5x+x=4-2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}4x=6\\6x=2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{1}{3}\end{cases}}\)
c. \(\left|2x-3\right|-\left|3x+2\right|=0\)
\(\Leftrightarrow\left|2x-3\right|=\left|3x+2\right|\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=3x+2\\2x-3=-3x-2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3x=3+2\\2x+3x=3-2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-x=5\\5x=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=\frac{1}{5}\end{cases}}\)
d, e tương tự
Ta có 3 x 2 + 8x + 5 = 0
ó 3 x 2 + 3x + 5x + 5 = 0 ó 3x(x + 1) + 5(x + 1) = 0
ó (3x + 5)(x + 1) = 0
Vậy x = - 5 3 ; x = - 1
Đáp án cần chọn là: A