\(\frac{-6}{30}=\frac{x}{-20}\)

nhân chéo  \(x\cdot30=\left(-6\right)\cdot\left(-20\right)\)

=>\(30x=120\)

\(x=4\)

\(\frac{-6}{30}=\frac{3}{y}\)

nhân chéo => \(-6x=90\)

\(x=-15\)

\(\frac{-6}{30}=\frac{z}{5}\)

nhân chéo => \(30z=-30\)

\(z=-1\)

x/-20 = -6/30 

=> 30x = 120 

<=> x = 4 

3/y = -6/30 

=> -6y = 90 

<=> y = -15 

z/5 = -6/30 

=> -6z = 150 

<=> z = - 25 

x=4,y=-15,z=1

-4/11+5/6+7/11-5/6

=(-4/11+7/11)+(5/6-5/6)

=3/11+0

=3/11

Làm là giúp đủ 3 câu hộ mình. 1 câu k k đâu !

3/7*(0,5+11/2)-2/7:25%

=3/7*(0,5+11/2)-2/7:1/4

=3/7*(0,5+11/2)-2/7*4

=3/7*(1/2+11/2)-2/7*4

=3/7*6-2/7*4

=18/7-8/7

=10/7

TL

 S= ( 1+ 3+ 3^2+ 3^3+ 3^4+ 3^5+ 3^6+ 3^7+ 3^8+ 3^9)

3.S=3.( 1+ 3+ 3^2+ 3^3+ 3^4+ 3^5+ 3^6+ 3^7+ 3^8+ 3^9)

3S=3+3^2+3^3+....+3^10

3S-S=3+3^2+3^3+....+3^10-(1+ 3+ 3^2+ 3^3+ 3^4+ 3^5+ 3^6+ 3^7+ 3^8+ 3^9)

2S=3^10-1

S=3^10-1/2

HỌC TỐT NHÉ

=17-5+9-18

=26-23

=3

Lời giải:
Theo bài ra ta có:

$3x=2y; 4y=5z$
$\Rightarrow \frac{x}{2}=\frac{y}{3}; \frac{y}{5}=\frac{z}{4}$

$\Rightarrow \frac{x}{10}=\frac{y}{15}=\frac{z}{12}$

Đặt $\frac{x}{10}=\frac{y}{15}=\frac{z}{12}=k$

$\Rightarrow x=10k; y=15k; z=12k$
Khi đó:

$3x^2-y^2+z^2=876$

$\Rightarrow 3(10k)^2-(15k)^2+(12k)^2=876$

$\Rightarrow 219k^2=876$

$\Rightarrow k^2=4$
$\Rightarrow k=\pm 2$

Nếu $k=2$ thì $x=10k=20; y=15k=30; z=12k=24$

Nếu $k=-2$ thì $x=10k=-20; y=15k=-30; z=12k=-24$

\(3x+xy-y=5\)

\(\Rightarrow3x+xy-y-3=5-3\)

\(\Rightarrow x\left(3+y\right)-\left(3+y\right)=2\)

\(\Rightarrow\left(y+3\right)\left(x-1\right)=2\)

Mà \(Ư_2=\left\{\pm1;\pm2\right\}\)

TH1 : \(\hept{\begin{cases}y+3=1\\x-1=2\end{cases}\Rightarrow\hept{\begin{cases}y=-2\\x=3\end{cases}}}\)

Th2 : \(\hept{\begin{cases}y+3=-1\\x-1=-2\end{cases}\Rightarrow\hept{\begin{cases}y=-4\\x=-1\end{cases}}}\)

TH3 : \(\hept{\begin{cases}y+3=2\\x-1=1\end{cases}\Rightarrow\hept{\begin{cases}y=-1\\x=2\end{cases}}}\)

TH4 : \(\hept{\begin{cases}y+3=-2\\x-1=-1\end{cases}\Rightarrow\hept{\begin{cases}y=-5\\x=0\end{cases}}}\)

KL...

Phạm Thị Thùy Linh cám ơn bạn lắm lắm...))

Đặt \(A=7^5+7^6+...+7^{100}\)

\(7A=7^6+7^7+...+7^{101}\\7A-A=(7^6+7^7+...+7^{101})-(7^5+7^6+...+7^{100})\\6A=7^{101}-7^5\\\Rightarrow A=\dfrac{7^{101}-7^5}{6}\)

A=67101−75​
 

=>\(\dfrac{3}{x-5}-\dfrac{y}{3}=\dfrac{1}{6}\)

=>\(\dfrac{9-y\left(x-5\right)}{3\left(x-5\right)}=\dfrac{1}{6}\)

=>9-y(x-5)=1/2(x-5)

=>(x-5)(1/2+y)=9

=>(x-5)(2y+1)=18

=>\(\left(x-5;2y+1\right)\in\left\{\left(18;1\right);\left(-18;-1\right);\left(2;9\right);\left(-2;-9\right);\left(6;3\right);\left(-6;-3\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(23;0\right);\left(-13;-1\right);\left(7;4\right);\left(3;-5\right);\left(11;1\right);\left(-1;-2\right)\right\}\)

a)y² = 7 => y = \(\sqrt{7}hoặc-\sqrt{7}\)

Nếu y = \(\sqrt{7}\) thì :

x²y³ = 5 . y² .y

x²y³ = 5.7.\(\sqrt{7}\) = 35\(\sqrt{7}\)

Nếu y = -\(\sqrt{7}\)  thì :

x²y³ = 5.7. (-\(\sqrt{7}\)) = -35\(\sqrt{7}\)

b) x²y² = 5.7 = 35

x⁶y⁶ = (x²y²)³ = 35³ = 42875

c) làm tương tự câu (a).  Chia x làm 2 trường hợp bằng căng 5 hoặc cặng 5 rồi thế vô tính nhé bạn!