\(-52+\frac{2}{3}x=-46\)
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= 10/3.x+ 67/4=-53/4
10/3.x=-53/4-67/4
10/3.x=-30
x=-30:10/3
x=-9
nhân chéo là đc:
3(x+2)=-4(x-5)
3x+6=-4x+20
3x+4x=20-6
7x =14
x =2
Vậy x=2
\(\frac{x+3}{-4}=-\frac{9}{x+3}\)
\(\Leftrightarrow\left(x+3\right)\left(x+3\right)=-4\cdot\left(-9\right)\)
\(\Leftrightarrow\left(x+3\right)^2=36\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x+3\right)^2=6^2\\\left(x+3\right)^2=\left(-6\right)^2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x+3=6\\x+3=-6\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-9\end{cases}}\)
Vậy ....
quy đồng
\(\left(x+3\right)^2=36\)
\(\left(x+3\right)^2-6^2=0\)
áp dụng định lí " \(a^2-b^2=\left(a+b\right)\left(a-b\right)\) ta được
\(\left(x+3-6\right)\left(x+3+6\right)=0\)
\(x=3,x=-9\)
Bài 2 :
a) (1) \(CaO+CO_2\rightarrow CaCO_3\)
(2) \(CaO+H_2O\rightarrow Ca\left(OH\right)_2\)
(3) \(Ca\left(OH\right)_2+CO_2\rightarrow CaCO_3+H_2O\)
(4) \(CaCO_3\underrightarrow{t^o}CaO+CO_2\)
(5) \(CaO+2HCl\rightarrow CaCl_2+H_2O\)
Chúc bạn học tốt
b)(1) \(S+O_2\underrightarrow{t^o}SO_2\)
(2) \(2SO_2+O_2\underrightarrow{t^o,V_2O_5}2SO_3\)
(3) \(SO_3+H_2O\rightarrow H_2SO_4\)
(4) \(H_2SO_4+CuO\rightarrow CuSO_4+H_2O\)
(5) \(CuSO_4+2NaOH\rightarrow Cu\left(OH\right)_2+Na_2SO_4\)
(6) \(Cu\left(OH\right)_2\underrightarrow{t^o}CuO+H_2O\)
(7) \(SO_2+2NaOH\rightarrow Na_2SO_3+H_2O\)
Chúc bạn học tốt
11)\(\dfrac{3x+1}{x-5}+\dfrac{2x}{x-5}=\dfrac{3x+2x+1}{x-5}=\dfrac{5x+1}{x-5}\)
12)\(\dfrac{4-x^2}{x-3}+\dfrac{2}{x^2-9}=\dfrac{4-x^2}{x-3}+\dfrac{2}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(4-x^2\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{2}{\left(x-3\right)\left(x+3\right)}=\dfrac{2+\left(2-x\right)\left(2+x\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)
13)
\(\dfrac{3}{4x-2}+\dfrac{2x}{4x^2-1}=\dfrac{3}{2\left(2x-1\right)}+\dfrac{2x}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{3\left(2x+1\right)}{2\left(2x-1\right)\left(2x+1\right)}+\dfrac{2.2x}{2\left(2x-1\right)\left(2x+1\right)}=\dfrac{6x+3+4x}{2\left(2x-1\right)\left(2x+1\right)}=\dfrac{10x+3}{2\left(2x-1\right)\left(2x+1\right)}\)
14)
\(\dfrac{2x+1}{2x-4}+\dfrac{5}{x^2-4}=\dfrac{2x+1}{2\left(x-2\right)}+\dfrac{5}{\left(x-2\right)\left(x+2\right)}=\dfrac{\left(2x+1\right)\left(x+2\right)}{2\left(x-2\right)\left(x+2\right)}+\dfrac{5.2}{2\left(x-2\right)\left(x+2\right)}=\dfrac{2x^2+5x+12}{2\left(x-2\right)\left(x+2\right)}\)
Phân tích đa thức thành nhân tử :
\(52.143-52.39-8.26\)
\(=52.143-52.39-52.4\)
\(=52.\left(143-39-4\right)\)
#~~ Hết~~#
\(-52+\frac{2}{3}x=-46\)
\(\frac{2}{3}x=-46+52\)
\(\frac{2}{3}x=6\)
\(x=6:\frac{2}{3}\)
\(x=9\)
x=9 nha bn
đúng nha
Happy new year!!