bai toán này kho lắm

tớ cũng đanh thắc mắc câu này

a) 3^x = 3^-12. 3^-15 . 3³²  =3⁵

x = 5

b) 2^x = 2⁹ .2^-30. 2⁹ = 2^-12

x = -12

c) 2^x = 2¹⁴ / 2⁹ = 2⁵

x = 5

\(/x-\frac{1}{2}/=\frac{1}{3}\\ =>\orbr{\begin{cases}x-\frac{1}{2}=\frac{1}{3}\\x-\frac{1}{2}=-\frac{1}{3}\end{cases}}\\ =>\orbr{\begin{cases}x=\frac{1}{3}+\frac{1}{2}\\x=-\frac{1}{3}+\frac{1}{2}\end{cases}}\\ =>\orbr{\begin{cases}x=\frac{5}{6}\\x=\frac{1}{6}\end{cases}}\)

\(a,|x-\frac{1}{2}|=\frac{1}{3}\)

\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{2}=\frac{1}{3}\\x-\frac{1}{2}=-\frac{1}{3}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{5}{6}\\x=\frac{1}{6}\end{cases}}}\)

\(b,\frac{14}{15}:\frac{9}{10}=x:\frac{3}{7}\)

\(\frac{28}{27}=x:\frac{3}{7}\)

\(x=\frac{4}{9}\)

\(c,\frac{6}{5}:8=\frac{2}{5}:x\)

\(\frac{3}{20}=\frac{2}{5}:x\)

\(x=\frac{8}{3}\)

\(d,3x-\frac{7}{8}=\frac{5}{2}\)

\(3x=\frac{27}{8}\)

\(x=\frac{9}{8}\)

Mọi người ơi, giúp em với ạ

a) \(\left(9^4.8+9^4.5\right):\left(9^2.\left(10-1\right)\right)\)

=\(9^4.13:9^3=13.9=117\)

b) 100-(75-25)=100-50=50

\(a,\left(2^3\cdot9^4+9^3\cdot45\right):\left(9^2\cdot10-9^2\right)\)

\(< =>\left(8\cdot6561+729\cdot45\right):\left(81\cdot10-81\right)\)

\(< =>\left(52488+32805\right):\left(810-81\right)\)

\(< =>85293:729\)

\(< =>117\)

\(b,100-\left[75-\left(7-2\right)^2\right]\)

\(< =>100-\left[75-\left(5\right)^2\right]\)

\(< =>100-\left[75-25\right]\)

\(< =>100-50\)

\(< =>50\)

\(c,\left(20\cdot2^4+12\cdot2^4\right)-\left(48\cdot2^2\right):8^2\)

\(< =>\left[2^4\cdot\left(20+12\right)\right]-\cdot\left(48\cdot4\right):8^2\)

\(< =>\left[2^4\cdot32\right]-192:8^2\)

\(< =>\left[16\cdot32\right]-192:64\)

\(< =>512-3\)

\(< =>509\)

a, 4¹⁰. 2³⁰=2²⁰.2³⁰=2⁵⁰

b,9²⁵.27⁴.81³= 3⁵⁰.3¹².3¹²=3⁷⁴

Tương tự các câu khác....

a, 4¹⁰.2³⁰ = (2²)¹⁰.2³⁰ = 2²⁰.2³⁰ = 2⁵⁰
b, 9²⁵.27⁴.81³ = (3²)²⁵.(3³)⁴.(3⁴)³ = 3⁵⁰.3¹².3¹² = 3⁷⁴
c, 25⁵⁰.125⁵ = (5²)⁵⁰.(5³)⁵ = 5¹⁰⁰.5¹⁵ = 5¹¹⁵
d, 64³.4⁸.16⁴ = (2⁶)³.(2²)⁸.(2⁴)⁴ = 2¹⁸.2¹⁶.2¹⁶ = 2⁵⁰
e, 3⁸ : 3⁶ = 3²
2¹⁰ : 8³ = 2¹⁰ : (2³)³ = 2¹⁰ : 2⁹ = 2
12⁷ : 6⁷ = (12 : 6)⁷ = 2⁷
@Dương Tuyết Mai

cái đầu tiên làm sao ra được

2x2+2=6

3x3-3=6

4 ko làm được

5:5+5=6

6x6:6=6

7-(7:7)=6

8,9,10 ko làm được

Lời giải:

a. ĐKXĐ: $x\geq -9$

PT $\Leftrightarrow x+9=7^2=49$

$\Leftrightarrow x=40$ (tm)

b. ĐKXĐ: $x\geq \frac{-3}{2}$

PT $\Leftrightarrow 4\sqrt{2x+3}-\sqrt{4(2x+3)}+\frac{1}{3}\sqrt{9(2x+3)}=15$

$\Leftrightarrow 4\sqrt{2x+3}-2\sqrt{2x+3}+\sqrt{2x+3}=15$

$\Leftrgihtarrow 3\sqrt{2x+3}=15$

$\Leftrightarrow \sqrt{2x+3}=5$

$\Leftrightarrow 2x+3=25$

$\Leftrightarrow x=11$ (tm)

c.

PT \(\Leftrightarrow \left\{\begin{matrix} 2x+1\geq 0\\ x^2-6x+9=(2x+1)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ 3x^2+10x-8=0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq \frac{-1}{2}\\ (3x-2)(x+4)=0\end{matrix}\right.\)

\(\Leftrightarrow x=\frac{2}{3}\)

d. ĐKXĐ: $x\geq 1$

PT \(\Leftrightarrow \sqrt{(x-1)+4\sqrt{x-1}+4}-\sqrt{(x-1)+6\sqrt{x-1}+9}=9\)

\(\Leftrightarrow \sqrt{(\sqrt{x-1}+2)^2}-\sqrt{(\sqrt{x-1}+3)^2}=9\)

\(\Leftrightarrow \sqrt{x-1}+2-(\sqrt{x-1}+3)=9\)

\(\Leftrightarrow -1=9\) (vô lý)

Vậy pt vô nghiệm.

a) \(\sqrt{x+9}=7\left(x\ge-9\right)\Rightarrow x+9=49\Rightarrow x=40\)

b) \(4\sqrt{2x+3}-\sqrt{8x+12}+\dfrac{1}{3}\sqrt{18x+27}=15\left(x\ge-\dfrac{3}{2}\right)\)

\(\Rightarrow4\sqrt{2x+3}-\sqrt{4\left(2x+3\right)}+\dfrac{1}{3}\sqrt{9\left(2x+3\right)}=15\)

\(\Rightarrow4\sqrt{2x+3}-2\sqrt{2x+3}+\sqrt{2x+3}=15\)

\(\Rightarrow3\sqrt{2x+3}=15\Rightarrow\sqrt{2x+3}=5\Rightarrow2x+3=25\Rightarrow x=11\)

c) \(\sqrt{x^2-6x+9}=2x+1\)

Vì \(VT\ge0\Rightarrow VP\ge0\Rightarrow x\ge-\dfrac{1}{2}\)

\(\Rightarrow\sqrt{\left(x-3\right)^2}=2x+1\Rightarrow\left|x-3\right|=2x+1\Rightarrow\left[{}\begin{matrix}x-3=2x+1\\x-3=-2x-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-4\left(l\right)\\x=\dfrac{2}{3}\end{matrix}\right.\)

d) \(\sqrt{x+3+4\sqrt{x-1}}-\sqrt{x+8+6\sqrt{x-1}}=9\left(x\ge1\right)\)

\(\Rightarrow\sqrt{x-1+4\sqrt{x-1}+4}-\sqrt{x-1+6\sqrt{x-1}+9}=9\)

\(\Rightarrow\sqrt{\left(\sqrt{x-1}+2\right)^2}-\sqrt{\left(\sqrt{x-1}+3\right)^2}=9\)

\(\Rightarrow\left|\sqrt{x-1}+2\right|-\left|\sqrt{x-1}+3\right|=9\)

\(\Rightarrow\sqrt{x-1}+2-\sqrt{x-1}-3=9\Rightarrow-1=9\) (vô lý)

Ta có:

a) ( 45 – 5 x 9 ) x 1 x 2 x 3 x 4 x 5 x 6 x 7

= (45 – 45) x 1 x 2 x 3 x 4 x 5 x 6 x 7

= 0 x 1 x 2 x 3 x 4 x 5 x 6 x 7

= 0

b) (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) x (72 – 8 x 8 – 8)

= (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) x (72 – 64 – 8)

= (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) x 0

= 0

c) (36 – 4 x 9) : (3 x 5 x 7 x 9 x 11)

= (36 – 36) : (3 x 5 x 7 x 9 x 11)

= 0 : (3 x 5 x 7 x 9 x 11)

= 0

d) (27 – 3 x 9) : 9 x 1 x 3 x 5 x 7

= (27 – 27) : 9 x 1 x 3 x 5 x 7

= 0 : 9 x 1 x 3 x 5 x 7

=0

a) ( 45 – 5 x 9 ) x 1 x 2 x 3 x 4 x 5 x 6 x 7

= 0 x 1 x 2 x 3 x 4 x 5 x 6 x 7

b) (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) x (72 – 8 x 8 – 8)

= (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) x 0

c) (36 – 4 x 9) : (3 x 5 x 7 x 9 x 11)

= 0 : (3 x 5 x 7 x 9 x 11)

d) (27 – 3 x 9) : 9 x 1 x 3 x 5 x 7

= 0 : 9 x 1 x 3 x 5 x 7                                                                                                                  Nếu đúng thì k cho mình nhé bạn!

a. ( 45 – 5 x 9 ) x 1 x 2 x 3 x 4 x 5 x 6 x 7

= (45 – 45) x 1 x 2 x 3 x 4 x 5 x 6 x 7

= 0 x 1 x 2 x 3 x 4 x 5 x 6 x 7

= 0

b. (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) x (72 – 8 x 8 – 8)

= (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) x (72 – 64 – 8)

= (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10) x 0

= 0

c. (36 – 4 x 9) : (3 x 5 x 7 x 9 x 11)

= (36 – 36) : (3 x 5 x 7 x 9 x 11)

= 0 : (3 x 5 x 7 x 9 x 11)

= 0

d. (27 – 3 x 9) : 9 x 1 x 3 x 5 x 7

= (27 – 27) : 9 x 1 x 3 x 5 x 7

= 0 : 9 x 1 x 3 x 5 x 7

=0