Đại số lớp 7

\(\left(a-2\right)^2=16\)

\(\Leftrightarrow\left(a-2\right)^2=\pm\sqrt{16}\)

\(\Leftrightarrow\left(a-2\right)=\pm4\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}a-2=4\\a-2=-4\end{array}\right.\Leftrightarrow\left[\begin{array}{nghiempt}a=6\\a=-2\end{array}\right.\)

Vậy \(a\in\left\{6;-2\right\}\)

giải:

ta có :

\(\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}}{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}}:\frac{3+\frac{3}{2}+\frac{3}{3}+\frac{3}{4}}{2-\frac{2}{2}+\frac{2}{3}-\frac{2}{4}}\)

\(\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}}{1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}}.\frac{2\left(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}{3\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)}=\frac{2}{3}\)

Đặt \(\frac{x}{4}=\frac{y}{7}=k\left(k\in Q,k\ne0\right)\)

\(\Rightarrow\begin{cases}x=4k\\y=7k\end{cases}\)

Theo đề: \(xy=4k.7k=112\)

\(\Rightarrow28k^2=112\)

\(\Rightarrow k^2=4\)

\(\Rightarrow k=2\)  hoặc \(k=-2\)

cho mk bổ sung nha

Với \(k=2\) thì:

\(x=4.2=8;y=7.2=14\)

Với \(k=-2\) thì:

\(x=4.\left(-2\right)=-8;y=7.\left(-2\right)=-14\)

Vậy \(\left(x,y\right)=\left(8;14\right)\),\(\left(-8;-14\right)\)

Đặt: \(\frac{x}{4}=\frac{y}{7}=k\)

=> \(\begin{cases}x=4k\\y=7k\end{cases}\)

Mà x.y = 112 => 4k.7k = 112 => 28k² =112 => k² = 4

=> k = 2 hoặc k = -2

Với k = 2 => \(\begin{cases}x=8\\y=14\end{cases}\)

Với k = -2 => \(\begin{cases}x=-8\\y=-14\end{cases}\)

Sai đề

a ) \(\frac{-2}{x}=\frac{x}{\frac{8}{25}}\)

\(\Leftrightarrow x^2=-2.-\frac{16}{25}\)

\(\Leftrightarrow x^2=\frac{16}{25}\) 

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\sqrt{\frac{16}{25}}\\x=-\sqrt{\frac{16}{25}}\end{array}\right.\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{4}{5}\\x=-\frac{4}{5}\end{array}\right.\)

Vậy ......................................

\(\left(2x-3\right)^2=81\)

\(\Leftrightarrow\left(2x-3\right)^2=\pm\sqrt{81}\)

\(\Leftrightarrow\left(2x-3\right)=\pm9\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}2x-3=9\\2x-3=-9\end{array}\right.\Leftrightarrow\left[\begin{array}{nghiempt}2x=12\\2x=-6\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=6\\x=-3\end{array}\right.\)

Vậy \(x\in\left\{6;-3\right\}\)

(2x- 3)²= 81

(2x- 3)²=9²

2x-3= 9 2x-3= -9

2x = 9+3 2x = (-9)+3

2x = 12 2x = -6

x = 12:2 x = -6:2

x = 6 x = -3

VAY x = 6 va x = -3

\(\frac{2x-1}{2x+3}=\frac{x-5}{x+1}\)

\(\Rightarrow\frac{2x-1}{2x+3}-\frac{x-5}{x+1}=0\)

\(\Rightarrow\frac{\left(2x-1\right)\left(x+1\right)}{\left(2x+3\right)\left(x+1\right)}-\frac{\left(2x+3\right)\left(x-5\right)}{\left(2x+3\right)\left(x+1\right)}=0\)

\(\Rightarrow\frac{2x^2+x-1-2x^2+7x+15}{\left(2x+3\right)\left(x+1\right)}=0\)

\(\Rightarrow\frac{8x+14}{\left(2x+3\right)\left(x+1\right)}=0\)

\(\Rightarrow8x+14=0\)

\(\Rightarrow8x=-14\)

\(\Rightarrow x=-\frac{7}{4}\)

cách đơn giản hơn thì nhân chéo 

\(10^x:5^y=20^y\)

\(=10^x:\left(2.5\right)^{y:2}=10^{y.2}\)

\(=10^x:10^{y:2}=10^{y.2}\)

\(=10^{x-y:2}=10^{y.2}\)

Phần còn lại bạn tự làm nha

\(\left(x-5\right)^2=\left(1-3x\right)^2\)

\(\left(x-5\right)^2:\left(1-3x\right)^2=0\)

\(\left(\frac{x-5}{1-3x}\right)^2=0\)

\(\Rightarrow\frac{x-5}{1-3x}=0\)

\(\Rightarrow x-5=0\)

\(\Rightarrow x=5\)

( x- 5)² = (1-3x)² 

(x - 5 )² : ( 1 - 3x )² = 0 

(\(\frac{x-5}{1-3x}\)) ² = 0

=>\(\frac{x-5}{1-3x}\) = 0 => x - 5 =0

     => x = 5