tìm x
3/x-1/ x 27 =234
x/9 -3/y = 1/18
X x(y-2) - y =3
mink cần gấp
K
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CM
22 tháng 2 2017
a) M = ( x – 1 ) 3 với x = 1001 thì M = 109.
b) N = ( x + y – 3 ) 3 với x = 2; y = 6 thì N = 125.
c) P = ( 3 xz 2 – 2 y ) 3 với x = 25; y = 150; z = 2 thì P = 0.
AH
Akai Haruma
Giáo viên
31 tháng 7 2021
Lời giải:
a. ĐKXĐ: $x\geq -9$
PT $\Leftrightarrow x+9=7^2=49$
$\Leftrightarrow x=40$ (tm)
b. ĐKXĐ: $x\geq \frac{-3}{2}$
PT $\Leftrightarrow 4\sqrt{2x+3}-\sqrt{4(2x+3)}+\frac{1}{3}\sqrt{9(2x+3)}=15$
$\Leftrightarrow 4\sqrt{2x+3}-2\sqrt{2x+3}+\sqrt{2x+3}=15$
$\Leftrgihtarrow 3\sqrt{2x+3}=15$
$\Leftrightarrow \sqrt{2x+3}=5$
$\Leftrightarrow 2x+3=25$
$\Leftrightarrow x=11$ (tm)
AH
Akai Haruma
Giáo viên
31 tháng 7 2021
c.
PT \(\Leftrightarrow \left\{\begin{matrix} 2x+1\geq 0\\ x^2-6x+9=(2x+1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ 3x^2+10x-8=0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ (3x-2)(x+4)=0\end{matrix}\right.\)
\(\Leftrightarrow x=\frac{2}{3}\)
d. ĐKXĐ: $x\geq 1$
PT \(\Leftrightarrow \sqrt{(x-1)+4\sqrt{x-1}+4}-\sqrt{(x-1)+6\sqrt{x-1}+9}=9\)
\(\Leftrightarrow \sqrt{(\sqrt{x-1}+2)^2}-\sqrt{(\sqrt{x-1}+3)^2}=9\)
\(\Leftrightarrow \sqrt{x-1}+2-(\sqrt{x-1}+3)=9\)
\(\Leftrightarrow -1=9\) (vô lý)
Vậy pt vô nghiệm.
1 tháng 11 2023
a: \(x^2-9-x^2\left(x^2-9\right)\)
\(=\left(x^2-9\right)-x^2\left(x^2-9\right)\)
\(=\left(x^2-9\right)\left(1-x^2\right)\)
\(=\left(1-x\right)\left(1+x\right)\left(x-3\right)\left(x+3\right)\)
b: \(x^2\left(x-y\right)+y^2\left(y-x\right)\)
\(=x^2\left(x-y\right)-y^2\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2-y^2\right)\)
\(=\left(x-y\right)\left(x-y\right)\left(x+y\right)=\left(x-y\right)^2\cdot\left(x+y\right)\)
c: \(x^3+27+\left(x+3\right)\left(x-9\right)\)
\(=\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)\)
\(=\left(x+3\right)\left(x^2-3x+9+x-9\right)\)
\(=\left(x+3\right)\left(x^2-2x\right)=x\left(x-2\right)\left(x+3\right)\)
d: \(x^2+5x+6\)
\(=x^2+2x+3x+6\)
\(=x\left(x+2\right)+3\left(x+2\right)=\left(x+2\right)\left(x+3\right)\)
e: \(3x^2-4x-4\)
\(=3x^2-6x+2x-4\)
\(=3x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(3x+2\right)\)
g: \(x^4+64y^4\)
\(=x^4+16x^2y^2+64y^4-16x^2y^2\)
\(=\left(x^2+8y^2\right)^2-\left(4xy\right)^2\)
\(=\left(x^2+8y^2-4xy\right)\left(x^2+8y^2+4xy\right)\)
1 tháng 11 2023
h: \(a^2+b^2+2a-2b-2ab\)
\(=a^2-2ab+b^2+2a-2b\)
\(=\left(a-b\right)^2+2\left(a-b\right)=\left(a-b\right)\left(a-b+2\right)\)
i: \(\left(x+1\right)^2-2\left(x+1\right)\left(y-3\right)+\left(y-3\right)^2\)
\(=\left(x+1-y+3\right)^2\)
\(=\left(x-y+4\right)^2\)
k: \(x^2\left(x+1\right)-2x\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-2x+1\right)\)
\(=\left(x+1\right)\left(x-1\right)^2\)
H
11 tháng 7 2021
3/5 - 1/3 x (2,48 + 0,52) x y : 60 : 5 = 1/5
1/3 x (2,48+0,52) x y : 60 : 5 = 2/5
1/3 x 3 x y : 60 : 5 = 2/5
y: 60 : 5 =2/5
y: 60 =2/5 x 5
y : 60 =2
y = 2 x 60
y =120
11 tháng 7 2021
Ta có: \(\dfrac{3}{5}-\dfrac{1}{3}\left(2.48+0.52\right)\cdot y:60:5=\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{3}{5}-\dfrac{1}{3}\cdot3\cdot y\cdot\dfrac{1}{60}\cdot\dfrac{1}{5}=\dfrac{1}{5}\)
\(\Leftrightarrow y\cdot\dfrac{1}{300}=\dfrac{2}{5}\)
hay y=120
HT
23 tháng 8 2016
ĐK: \(x\ge0;y\ge\frac{9}{2}\)
(1) \(\Leftrightarrow6\left(x+\frac{1}{2}\right)\sqrt{\left[3\left(x+\frac{1}{2}\right)\right]^2+\frac{27}{4}}=2y\sqrt{y^2+\frac{27}{4}}\)
Xét \(f\left(t\right)=2t\sqrt{t^2+\frac{27}{4}}\left(t>0\right)\)
\(f'\left(t\right)=2\sqrt{t^2+\frac{27}{4}}+\frac{2t^2}{\sqrt{t^2+\frac{27}{4}}}>0;\forall t>0\)
→ hàm đồng biến trên (0;+∞)
Mà \(f\left(3\left(x+\frac{1}{2}\right)\right)=f\left(y\right)\Leftrightarrow3\left(x+\frac{1}{2}\right)=y\)
Thế vào (2) ta được:
\(\left(6y+6\right)^2=24\sqrt{x}\left(6y-6\right)\Leftrightarrow\left(x+1\right)^2=4\sqrt{x}\left(x-1\right)\)
\(\Leftrightarrow\left(\sqrt{x}\right)^4-4\left(\sqrt{x}\right)^3+2\left(\sqrt{x}\right)^2+4\sqrt{x}+1=0\)
\(\Leftrightarrow\left(\sqrt{x}\right)^4+4\sqrt{x}+1-2\cdot x\cdot2\sqrt{x}-2\cdot x\cdot1+2\cdot1\cdot2\sqrt{x}=0\)
\(\Leftrightarrow\left(x-2\sqrt{x}-1\right)^2=0\)
\(\Leftrightarrow x-2\sqrt{x}-1=0\Leftrightarrow\sqrt{x}=1+\sqrt{2}\Leftrightarrow x=3+2\sqrt{2}\)
\(\Rightarrow y=\frac{21+12\sqrt{2}}{2}\)
1, http://123link.pro/dEkYZYmy