thu gọn 7^3*7^5

Oh no nhiều kí tự đặc biệt quá

dễ quá mình ko làm  đc 

1) ab=2 (I); bc=3 (II); ca=54 (III)

Lấy (I).(II).(III) ⇒ a² . b² . c² = 324 ⇒ abc = ±18

(II) ⇒ a= ±6 ; (I) ⇒ b= ±1/3 ; (II) ⇒ c= ±9

2) ab=5/3 (I); bc=4/5 (II); ca=3/4 (III)

Lấy (I).(II).(III) ⇒ a² . b² . c² = 1 ⇒ abc = ±1

(II) ⇒ a= ±5/4 ; (I) ⇒ b= ±4/3 ; (II) ⇒ c= ±3/5

3) a(a+b+c)= -12 (I)

    b(a+b+c)= 18 (II)

    c(a+b+c)= 30 (III)

Lấy (I)+(II)+(III) ⇒ (a+b+c)² = 36 ⇒ a+b+c = ±6

TH1 : a=6 ⇒ a= -12/6 = -2 ; b= 18/6 = 3 ; c= 30/6 = 5

TH2 : a=-6 ⇒ a= -12/-6 = 2 ; b= 18/-6 = -3 ; c= 30/-6 = -5

Lời giải:

Đặt $a-\frac{b}{2}=x; \frac{a}{2}-b=y$ thì $45^0< x< 180^0; -45^0< y< 90^0$

$\cos x=\frac{-1}{4}; 45^0< x< 180^0$ nên $\sin x=\frac{\sqrt{15}}{4}$

$\sin y=\frac{1}{3}; -45^0< y< 90^0$ nên $\cos y=\frac{2\sqrt{2}}{3}$

\(P=72\cos (2x-2y)+49=72[2\cos ^2(x-y)-1]+49=144\cos ^2(x-y)-23\)

\(=144(\cos x\cos y+\sin x\sin y)^2-23=-4\sqrt{30}\)

Đáp án C.

a) ta có :     y + 30/100y = -1,3

           hay       y(1+ 3/10)    = - 13/10

                       y.13/10 = -13/10

                       y= -13/10:13/10

                       y= -1

b)       y - 25/100y=1/2

hay    y(1-1/4)=1/2

         y.3/4 = 1/2

         y=2/3

c)    10/3y + 16,75 = -13,25

       10/3y= -13,25 - 16,75

       10/3y= -30

             y= - 30:10/3

             y= -9

2)

a) \(2\left|2x-3\right|=1\)

=> \(\left|2x-3\right|=1:2\)

=> \(\left|2x-3\right|=\frac{1}{2}\)

=> \(\left[{}\begin{matrix}2x-3=\frac{1}{2}\\2x-3=-\frac{1}{2}\end{matrix}\right.\) => \(\left[{}\begin{matrix}2x=\frac{1}{2}+3=\frac{7}{2}\\2x=\left(-\frac{1}{2}\right)+3=\frac{5}{2}\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=\frac{7}{2}:2\\x=\frac{5}{2}:2\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{7}{4}\\x=\frac{5}{4}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{7}{4};\frac{5}{4}\right\}.\)

b) \(7,5-3\left|5-2x\right|=-4,5\)

=> \(4,5\left|5x-2\right|=-4,5\)

=> \(\left|5x-2\right|=\left(-4,5\right):4,5\)

=> \(\left|5x-2\right|=-1\)

Ta luôn có: \(\left|x\right|>0\forall x\)

=> \(\left|5x-2\right|>-1\)

=> \(\left|5x-2\right|\ne-1\)

Vậy không tồn tại giá trị nào của \(x\) thỏa mãn yêu cầu đề bài.

c) \(\left|3x-4\right|+\left|3y+5\right|=0\)

Ta có: \(\left|3x-4\right|>\) hoặc \(=0\forall x\)

\(\left|3y+5\right|>\) hoặc \(=0\forall y.\)

=> \(\left|3x-4\right|+\left|3y+5\right|=0\)

=> \(\left[{}\begin{matrix}3x-4=0\\3y+5=0\end{matrix}\right.\) => \(\left[{}\begin{matrix}3x=0+4=4\\3y=0-5=-5\end{matrix}\right.\) => \(\left[{}\begin{matrix}x=4:3\\y=\left(-5\right):3\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{4}{3}\\y=-\frac{5}{3}\end{matrix}\right.\)

Vậy \(x\in\left\{\frac{4}{3}\right\};y\in\left\{-\frac{5}{3}\right\}.\)

Chúc bạn học tốt!

Bài 1:

a) \(-15,5.20,8+3,5.9,2-15,5.9,2+3,5.20,8\)

\(=20,8.\left(-15,5+3,5\right)+9,2.\left(-15,5+3,5\right)\)

\(=\left(-15,5+3,5\right).\left(20,8+9,2\right)\)

\(=\left(-12\right).30=-360\)

b) \(\left[\left(-19,95\right)+\left(-45,75\right)\right]+\left[4,95+5,75\right]\)

\(=\left[\left(-19,95\right)+4,95\right]+\left[\left(-45,75\right)+5,75\right]\)

\(=-15+\left(-40\right)=-55\)

Bài 2 :

\(a,2.\left|2x-3\right|=1\)

\(\Leftrightarrow\left|2x-3\right|=\frac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=\frac{1}{2}\\2x-3=-\frac{1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{1}{2}+3\\2x=-\frac{1}{2}+3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=\frac{7}{2}\\2x=\frac{5}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{4}\\x=\frac{5}{4}\end{matrix}\right.\)

Vậy : \(x\in\left\{\frac{7}{4},\frac{5}{4}\right\}\)

\(b,7.5-3\left|5-2x\right|=-4.5\)

\(\Leftrightarrow3.\left|5-2x\right|=7.5-\left(-4.5\right)=12\)

\(\Leftrightarrow\left|5-2x\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}5-2x=4\\5-2x=-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\2x=9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=\frac{9}{2}\end{matrix}\right.\)

Vậy : \(x\in\left\{\frac{1}{2},\frac{9}{2}\right\}\)

\(c,\left|3x-4\right|+\left|3y+5\right|=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x-4=0\\3y+5=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}3x=4\\3y=-5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{4}{3}\\y=-\frac{5}{3}\end{matrix}\right.\)

Vậy : \(\left(x,y\right)=\left(\frac{4}{3},-\frac{5}{3}\right)\)

Bài 3 :

a) \(2^{300}\) và \(3^{200}\)

Ta có : \(2^{300}=\left(2^3\right)^{100}=8^{100}\)

\(3^{200}=\left(3^2\right)^{100}=9^{100}\)

mà : \(9^{100}>8^{100}\Rightarrow3^{200}>2^{300}\)

Vậy : \(3^{200}>2^{300}\)

b) \(2^{30}+3^{30}+4^{30}\) và \(3.2.4^{10}\)

Ta có : \(3.2.4^{10}=6.\left(2^2\right)^{10}=6.2^{20}=3.2^{21}\)

Ta thấy : \(2^{30}>3.2^{21}\Rightarrow2^{30}+3^{30}+4^{30}>3.2^{21}\)

hay : \(2^{30}+3^{30}+4^{30}>3.2.4^{10}\)

Vậy : \(2^{30}+3^{30}+4^{30}>3.2.4^{10}\)

Chúc bạn học tốt !