giải giúp mk vs , mk đg cần gấp ...
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HD
3
14 tháng 12 2022
`(8x^3y^4z) : (4x^2y^4)`
`= (8:4) . (x^3 : x^2) . (y^4 : y^4) . z`
`= 2xz.`
PM
3
30 tháng 8 2021
PTHH: 2Al + 6HCl → 2AlCl3 + 3H2
PTHH: Mg(OH)2 + 2HCl → MgCl2 + 2H2O
PTHH: Fe2O3 + 6HCl → 2FeCl3 + 3H2O
PTHH: K2O + 2HCl → 2KCl + H2O
30 tháng 8 2021
Lần sau em đăng câu hỏi đúng box để mọi người hỗ trợ nha
B
22 tháng 8 2021
a)
=15(can6-1)/(6-1)+4/(4-2)-12/(3-4)
=3(can6-1)+2+12
=3\(\sqrt{6}\)-3+2+12
=17+3can6
các câu còn lại tương tự liên hợp mẫu
NL
Nguyễn Lê Phước Thịnh
CTVHS
6 tháng 6
VD1:
a: \(5\cdot\sqrt{25a^2}-25a\)
\(=5\cdot\left|5a\right|-25a\)
=-25a-25a(a<0)
=-50a
b: \(\sqrt{49a^2}+3a\)
\(=\sqrt{\left(7a\right)^2}+3a\)
=7a+3a
=10a
c: Đặt A=\(\sqrt{64a^2}-8a\)
\(=\sqrt{\left(8a\right)^2}-8a\)
=8|a|-8a
TH1: a>=0
=>A=8a-8a=0
TH2: a<0
=>A=-8a-8a=-16a
d: Đặt \(A=\sqrt{9a^6}-3a^3\)
\(=3\cdot\sqrt{a^6}-3a^3\)
\(=3\cdot\left|a^3\right|-3a^3\)
TH1: a>=0
=>\(A=3a^3-3a^3=0\)
TH2: a<0
=>\(A=-3a^3-3a^3=-6a^3\)
VD2:
a: \(4x-\sqrt{x^2-4x+4}\)
\(=4x-\sqrt{\left(x-2\right)^2}\)
=4x-|x-2|
=4x-(x-2)(x>=2)
=4x-x+2
=3x+2
b: \(3x+\sqrt{x^2+6x+9}\)
\(=3x+\sqrt{\left(x+3\right)^2}\)
=3x+|x+3|
=3x+(-x-3)(x<-3)
=2x-3
c: \(\frac{x+6\sqrt{x}+9}{x-9}\)
\(=\frac{\left(\sqrt{x}+3\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+3}{\sqrt{x}-3}\)
d: \(\frac{\sqrt{x^2+4x+4}}{x+2}\)
\(=\frac{\sqrt{\left(x+2\right)^2}}{x+2}\)
\(=\frac{\left|x+2\right|}{x+2}=\pm1\)
KK
1
NN
4
8 tháng 12 2019
+) nếu x; y đều dương => x + y = 10
+) nếu x; y đều âm => - x + (- y) = 10 <=> x + y = -10
hc tốt
8 tháng 12 2019
Xét 2 TH
TH1: /x/+/y/= 10
\(\Rightarrow\)x+y=10
TH2 : /x/+/y/=10
\(\Rightarrow\)-x + (-y) =10
\(\Rightarrow\)-x - -y = 10
=> - ( x+y ) = 10
=> x+y = -10
Vậy: x+y= -10 hoặc 10
HP
6 tháng 9 2021
4.
a, \(A=\sqrt[3]{15\sqrt{3}+26}=\sqrt[3]{\left(\sqrt{3}+2\right)^3}=\sqrt{3}+2\)
b, \(B=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)
\(\Rightarrow2B=\sqrt[3]{40+16\sqrt{13}}+\sqrt[3]{40-16\sqrt{13}}\)
\(=\sqrt[3]{\left(\sqrt{13}+1\right)^3}+\sqrt[3]{\left(\sqrt{13}-1\right)^3}\)
\(=\sqrt{13}+1+\sqrt{13}-1=2\sqrt{13}\)
\(\Rightarrow B=\sqrt{13}\)
c, \(C=\sqrt[3]{182-\sqrt{33125}}+\sqrt[3]{182+\sqrt{33125}}\)
\(\Rightarrow C^3=364+3\sqrt[3]{182-\sqrt{33125}}.\sqrt[3]{182+\sqrt{33125}}\left(\sqrt[3]{182-\sqrt{33125}}+\sqrt[3]{182+\sqrt{33125}}\right)\)
\(=364-3C\)
\(\Rightarrow C^3+3C-364=0\)
\(\Leftrightarrow C=7\)
PT
1
a) \(-1\dfrac{4}{5}x-1\dfrac{1}{10}=25\%\)
\(\Leftrightarrow\dfrac{-9}{5}x-\dfrac{11}{10}=\dfrac{1}{4}\)
\(\Leftrightarrow x=\dfrac{-3}{4}\)
b) \(\left(2x+\dfrac{3}{5}\right)^2-\dfrac{9}{25}=0\)
\(\Leftrightarrow\left(2x+\dfrac{3}{5}\right)^2-\left(\dfrac{3}{5}\right)^2=0\)
\(\Leftrightarrow\left(2x+\dfrac{3}{5}+\dfrac{3}{5}\right)\left(2x+\dfrac{3}{5}-\dfrac{3}{5}\right)=0\)
\(\Leftrightarrow2x\left(2x+\dfrac{6}{5}\right)=0\)
\(\Leftrightarrow4x^2+\dfrac{12}{5}=0\)
\(\Leftrightarrow x^2=\dfrac{-3}{5}\)
\(\Rightarrow x\in\varnothing\)
c) \(1,25-\left|\dfrac{x}{3}+2\right|=\left(-\dfrac{1}{2}\right)^3\)
\(\Leftrightarrow\dfrac{5}{4}-\left|\dfrac{x}{3}+2\right|=\dfrac{-1}{8}\)
\(\Leftrightarrow\dfrac{11}{8}-\left|\dfrac{x}{3}+2\right|=0\)
\(\left|\dfrac{x}{3}+2\right|=\dfrac{x}{3}+2\) khi \(\dfrac{x}{3}+2\ge0\Rightarrow x\ge-6\)
Với \(x\ge-6\) ta có: \(\dfrac{11}{8}-\dfrac{x}{3}-2=0\)
\(\Rightarrow x=\dfrac{-15}{8}\left(TM\right)\)
\(\left|\dfrac{x}{3}+2\right|=-\dfrac{x}{3}-2\) khi \(\dfrac{x}{3}+2< 0\Rightarrow x< -6\)
Với \(x< -6\) ta có: \(\dfrac{11}{8}+\dfrac{x}{3}+2=0\)
\(\Rightarrow x=-\dfrac{81}{8}\) (TM)
Vậy..
d) \(\dfrac{5}{6}\left(4x-\dfrac{2}{5}\right)-3\left(\dfrac{1}{12}x-\dfrac{1}{2}\right)=2x-\dfrac{5}{9}\)
\(\Leftrightarrow\dfrac{10}{3}x-\dfrac{1}{3}-\dfrac{1}{4}x+\dfrac{3}{2}=\dfrac{18x-5}{9}\)
\(\Leftrightarrow\dfrac{120x-12-9x+54-72x+20}{36}=0\)
\(\Rightarrow39x+62=0\)
\(\Rightarrow x=\dfrac{-62}{39}\)
e) \(\left|7x-4\right|-5x=28\)
\(\left|7x-4\right|=7x-4\) khi \(7x-4\ge0\Rightarrow x\ge\dfrac{4}{7}\)
Với \(x\ge\dfrac{4}{7}\) ta có: \(7x-4-5x=28\)
\(\Rightarrow x=16\left(TM\right)\)
\(\left|7x-4\right|=4-7x\) khi \(7x-4< 0\Rightarrow x< \dfrac{4}{7}\)
Với \(x< \dfrac{4}{7}\) ta có: \(4-7x-5x=28\)
\(\Rightarrow x=-2\left(TM\right)\)
Vậy...
d) \(\dfrac{5}{6}\left(4x-\dfrac{2}{5}\right)-3\left(\dfrac{1}{12}x-\dfrac{1}{2}\right)=2x-\dfrac{5}{9}\)
\(\Rightarrow\dfrac{10}{3}x-\dfrac{1}{3}-\dfrac{1}{4}x+\dfrac{3}{2}=2x-\dfrac{5}{9}\)
\(\Rightarrow\dfrac{10}{3}x-\dfrac{1}{4}x-2x=\dfrac{1}{3}-\dfrac{3}{2}-\dfrac{5}{9}\)
\(\Rightarrow\left(\dfrac{10}{3}-\dfrac{1}{4}-2\right)x=\dfrac{-31}{18}\)
\(\Rightarrow\dfrac{13}{12}x=\dfrac{-31}{18}\)
\(\Rightarrow x=\dfrac{-62}{39}.\)
Vậy \(x=\dfrac{-62}{39}.\)
e) \(\left|7x-4\right|-5x=28\)
\(\Rightarrow\left|7x-4\right|=5x+28\)
+) \(TH1:7x-4\ge0\Rightarrow7x\ge4\Rightarrow x\ge\dfrac{4}{7}\)
\(7x-4=5x+28\)
\(\Rightarrow7x-5x=4+28\)
\(\Rightarrow2x=32\)
\(\Rightarrow x=16\) (nhận)
+) \(TH2:7x-4< 0\Rightarrow7x< 4\Rightarrow x< \dfrac{4}{7}\)
\(-7x+4=5x+28\)
\(\Rightarrow-7x-5x=-4+28\)
\(\Rightarrow-12x=24\)
\(\Rightarrow x=-2\) (nhận)
Vậy \(x\in\left\{16;-2\right\}\).
Mấy câu kia dễ tự làm.