Câu 1 : 

a, \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}=\frac{2x-1}{3}-\frac{3-x}{4}\)

\(\Leftrightarrow\frac{6x+3}{4}+\frac{3-x}{4}=\frac{2x-1}{3}+\frac{5x+3}{6}\)

\(\Leftrightarrow\frac{5x+6}{4}=\frac{9x+1}{6}\Leftrightarrow\frac{30x+36}{24}=\frac{36x+4}{24}\)

Khử mẫu : \(30x+36=36x+4\Leftrightarrow-6x=-32\Leftrightarrow x=\frac{32}{6}=\frac{16}{3}\)

tương tự 

\(\frac{19}{4}-\frac{2\left(3x-5\right)}{5}=\frac{3-2x}{10}-\frac{3x-1}{4}\)

\(< =>\frac{19.5}{20}-\frac{8\left(3x-5\right)}{20}=\frac{2\left(3-2x\right)}{20}-\frac{5\left(3x-1\right)}{20}\)

\(< =>95-24x+40=6-4x-15x+5\)

\(< =>-24x+135=-19x+11\)

\(< =>5x=135-11=124\)

\(< =>x=\frac{124}{5}\)

\(\frac{\left(x-2\right).3}{2}+3+\frac{\left(x-3\right).5}{3}+5+\frac{\left(x-5\right).2}{5}+2=10\)

\(< =>\frac{\left(x-2\right).3.15}{30}+\frac{\left(x-3\right).5.10}{30}+\frac{\left(x-5\right).2.6}{30}=10-2-3-5\)

\(< =>\frac{\left(x-2\right).45+\left(x-3\right).50+\left(x-5\right).12}{30}=0\)

\(< =>45x-90+50x-150+12x-60=0\)

\(< =>107x-300=0< =>x=\frac{300}{107}\)

1: \(2^x=64\)

=>\(x=log_264=6\)

2: \(2^x\cdot3^x\cdot5^x=7\)

=>\(\left(2\cdot3\cdot5\right)^x=7\)

=>\(30^x=7\)

=>\(x=log_{30}7\)

3: \(4^x+2\cdot2^x-3=0\)

=>\(\left(2^x\right)^2+2\cdot2^x-3=0\)

=>\(\left(2^x\right)^2+3\cdot2^x-2^x-3=0\)

=>\(\left(2^x+3\right)\left(2^x-1\right)=0\)

=>\(2^x-1=0\)

=>\(2^x=1\)

=>x=0

4: \(9^x-4\cdot3^x+3=0\)

=>\(\left(3^x\right)^2-4\cdot3^x+3=0\)

Đặt \(a=3^x\left(a>0\right)\)

Phương trình sẽ trở thành:

\(a^2-4a+3=0\)

=>(a-1)(a-3)=0

=>\(\left[{}\begin{matrix}a-1=0\\a-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=1\left(nhận\right)\\a=3\left(nhận\right)\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}3^x=1\\3^x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\)

5: \(3^{2\left(x+1\right)}+3^{x+1}=6\)

=>\(\left[3^{x+1}\right]^2+3^{x+1}-6=0\)

=>\(\left(3^{x+1}\right)^2+3\cdot3^{x+1}-2\cdot3^{x+1}-6=0\)

=>\(3^{x+1}\left(3^{x+1}+3\right)-2\left(3^{x+1}+3\right)=0\)

=>\(\left(3^{x+1}+3\right)\left(3^{x+1}-2\right)=0\)

=>\(3^{x+1}-2=0\)

=>\(3^{x+1}=2\)

=>\(x+1=log_32\)

=>\(x=-1+log_32\)

6: \(\left(2-\sqrt{3}\right)^x+\left(2+\sqrt{3}\right)^x=2\)
=>\(\left(\dfrac{1}{2+\sqrt{3}}\right)^x+\left(2+\sqrt{3}\right)^x=2\) 

=>\(\dfrac{1}{\left(2+\sqrt{3}\right)^x}+\left(2+\sqrt{3}\right)^x=2\)

Đặt \(b=\left(2+\sqrt{3}\right)^x\left(b>0\right)\)

Phương trình sẽ trở thành:

\(\dfrac{1}{b}+b=2\)

=>\(b^2+1=2b\)

=>\(b^2-2b+1=0\)

=>(b-1)²=0

=>b-1=0

=>b=1

=>\(\left(2+\sqrt{3}\right)^x=1\)

=>x=0

7: ĐKXĐ: \(x^2+3x>0\)

=>x(x+3)>0

=>\(\left[{}\begin{matrix}x>0\\x< -3\end{matrix}\right.\)
\(log_4\left(x^2+3x\right)=1\)

=>\(x^2+3x=4^1=4\)

=>\(x^2+3x-4=0\)

=>(x+4)(x-1)=0

=>\(\left[{}\begin{matrix}x+4=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)

\(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)

\(\Rightarrow\frac{x+1}{9}+1+\frac{x+2}{8}+1=\frac{x+3}{7}+1+\frac{x+4}{6}+1\)

\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}=\frac{x+10}{7}+\frac{x+10}{6}\)

\(\Rightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)

mà \(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\ne0\)

=> x + 10 = 0 => x = -10

                                                                         Vậy x = -10

\(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)

\(\Leftrightarrow\left(\frac{x+1}{9}+1\right)+\left(\frac{x+2}{8}+1\right)=\left(\frac{x+3}{7}+1\right)+\left(\frac{x+4}{6}+1\right)\)

\(\Leftrightarrow\frac{x+10}{9}+\frac{x+10}{8}=\frac{x+10}{7}+\frac{x+10}{6}\)

\(\Leftrightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)

\(\Leftrightarrow x+10=0\Leftrightarrow x=-10\)

a) \(6\sqrt{x-1}-\dfrac{1}{3}\cdot\sqrt{9x-9}+\dfrac{7}{2}\sqrt{4x-4}=24\) (ĐK: \(x\ge1\)) 

\(\Leftrightarrow6\sqrt{x-1}-\dfrac{1}{3}\cdot\sqrt{9\left(x-1\right)}+\dfrac{7}{2}\sqrt{4\left(x-1\right)}=24\)

\(\Leftrightarrow6\sqrt{x-1}-\dfrac{1}{3}\cdot3\sqrt{x-1}+\dfrac{7}{2}\cdot2\sqrt{x-1}=24\)

\(\Leftrightarrow6\sqrt{x-1}-\sqrt{x-1}+7\sqrt{x-1}=24\)

\(\Leftrightarrow12\sqrt{x-1}=24\)

\(\Leftrightarrow\sqrt{x-1}=\dfrac{24}{12}\)

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

\(\Leftrightarrow x-1=4\)

\(\Leftrightarrow x=4+1\)

\(\Leftrightarrow x=5\left(tm\right)\)

b) \(\dfrac{1}{2}\sqrt{4x+8}-2\sqrt{x+2}-\dfrac{3}{7}\sqrt{49x+98}=-8\) (ĐK: \(x\ge-2\))

\(\Leftrightarrow\dfrac{1}{2}\cdot2\sqrt{x+2}-2\sqrt{x+2}-\dfrac{3}{7}\cdot7\sqrt{x+2}=-8\)

\(\Leftrightarrow\sqrt{x+2}-2\sqrt{x+2}-3\sqrt{x+2}=-8\)

\(\Leftrightarrow-4\sqrt{x+2}=-8\)

\(\Leftrightarrow\sqrt{x+2}=\dfrac{-8}{-4}\)

\(\Leftrightarrow\sqrt{x+2}=2\)

\(\Leftrightarrow x+2=4\)

\(\Leftrightarrow x=4-2\)

\(\Leftrightarrow x=2\left(tm\right)\)

\(\Rightarrow\left(\frac{x+1}{9}+1\right)+\left(\frac{x+2}{8}+1\right)=\left(\frac{x+3}{7}+1\right)+\left(\frac{x+4}{6}+1\right)\)

\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}=\frac{x+10}{7}+\frac{x+10}{6}\)

\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)

\(\Rightarrow\left(x+10\right).\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)

\(\Rightarrow x+10=0\Rightarrow x=-10\)

a) 7x - 35 = 0

<=> 7x = 0 + 35

<=> 7x = 35

<=> x = 5

b) 4x - x - 18 = 0

<=> 3x - 18 = 0

<=> 3x = 0 + 18

<=> 3x = 18

<=> x = 5

c) x - 6 = 8 - x

<=> x - 6 + x = 8

<=> 2x - 6 = 8

<=> 2x = 8 + 6

<=> 2x = 14

<=> x = 7

d) 48 - 5x = 39 - 2x

<=> 48 - 5x + 2x = 39

<=> 48 - 3x = 39

<=> -3x = 39 - 48

<=> -3x = -9

<=> x = 3

có bị viết nhầm thì thông cảm nha!

la`thu'hai nga`y 19 nhe

Câu 2/

Điều kiện xác định b tự làm nhé:

\(\frac{6}{x^2-9}+\frac{4}{x^2-11}-\frac{7}{x^2-8}-\frac{3}{x^2-12}=0\)

\(\Leftrightarrow x^4-25x^2+150=0\)

\(\Leftrightarrow\left(x^2-10\right)\left(x^2-15\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2=10\\x^2=15\end{cases}}\)

Tới đây b làm tiếp nhé.

a. ĐK: \(\frac{2x-1}{y+2}\ge0\)

Áp dụng bđt Cô-si ta có: \(\sqrt{\frac{y+2}{2x-1}}+\sqrt{\frac{2x-1}{y+2}}\ge2\)

\(\)Dấu bằng xảy ra khi  \(\frac{y+2}{2x-1}=1\Rightarrow y+2=2x-1\Rightarrow y=2x-3\) 

Kết hợp với pt (1) ta tìm được x = -1, y = -5 (tmđk)

b. \(pt\Leftrightarrow\left(\frac{6}{x^2-9}-1\right)+\left(\frac{4}{x^2-11}-1\right)-\left(\frac{7}{x^2-8}-1\right)-\left(\frac{3}{x^2-12}-1\right)=0\)

\(\Leftrightarrow\left(15-x^2\right)\left(\frac{1}{x^2-9}+\frac{1}{x^2-11}+\frac{1}{x^2-8}+\frac{1}{x^2-12}\right)=0\)

\(\Leftrightarrow x^2-15=0\Leftrightarrow\orbr{\begin{cases}x=\sqrt{15}\\x=-\sqrt{15}\end{cases}}\)

BALABOLO

TK NHA

Bạn cần viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) đẻ được hỗ trợ tốt hơn. Viết như thế kia rất khó đọc => khả năng bị bỏ qua bài cao.

a: =>3x=3

=>x=1

b: =>12x-2(5x-1)=3(8-3x)

=>12x-10x+2=24-9x

=>2x+2=24-9x

=>11x=22

=>x=2

c: =>2x-3(2x+1)=x-6x

=>-5x=2x-6x-3=-4x-3

=>-x=-3

=>x=3

d: =>2x-5=0 hoặc x+3=0

=>x=5/2 hoặc x=-3

e: =>x+2=0

=>x=-2

\(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)

\(\Leftrightarrow\frac{x+1}{9}+1+\frac{x+2}{8}+1=\frac{x+3}{7}+1+\frac{x+4}{6}+1\)

\(\Leftrightarrow\frac{x+10}{9}+\frac{x+10}{8}=\frac{x+10}{7}+\frac{x+10}{6}\)

\(\Leftrightarrow\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)

\(\Leftrightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)

Mà \(\frac{1}{9}< \frac{1}{8}< \frac{1}{7}< \frac{1}{6}\)nên \(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}< 0\)

Suy ra x + 10 = 0

Vậy x = -10

Pt ban đầu tương đương :

\(\left(\frac{x+1}{9}+1\right)+\left(\frac{x+2}{8}+1\right)=\left(\frac{x+3}{7}+1\right)+\left(\frac{x+4}{6}+1\right)\)

\(\Leftrightarrow\frac{x+10}{9}+\frac{x+10}{8}=\frac{x+10}{7}+\frac{x+10}{6}\)

\(\Leftrightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)

Mà : \(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\ne0\)

\(\Rightarrow x+10=0\)

\(\Leftrightarrow x=-10\) ( thỏa mãn )

Vậy pt đã cho có tập nghiệm \(S=\left\{-10\right\}\)

a.$\frac{x+1}{9}\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}$

=>$\frac{x+1}{9}+1+\frac{x+2}{8}+1=\frac{x+3}{7}+1+\frac{x+4}{6}+1$

<=> $\frac{x+1+9}{9}+\frac{x+2+8}{8}=\frac{x+3+7}{7}+\frac{x+4+6}{6}$

<=>$\frac{x+10}{9}+\frac{x+10}{8}=\frac{x+10}{7}+\frac{x+10}{6}$

<=> $\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0$

<=> $\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0$

<=> x+10=0

<=> x=-10

Vậy tập nghiệm của phương trình trên là S=$\left\{-10\right\}$

Vậy phương trình có nghiệm duy nhất là x = -10.

\(\dfrac{x+1}{9}+\dfrac{x+2}{8}=\dfrac{x+3}{7}+\dfrac{x+4}{6}\\ \Leftrightarrow\dfrac{x+1}{9}+1+\dfrac{x+2}{8}+1=\dfrac{x+3}{7}+1+\dfrac{x+4}{6}+1\\ \Leftrightarrow\dfrac{x+10}{9}+\dfrac{x+10}{8}=\dfrac{x+10}{7}+\dfrac{x+10}{6}\\ \Leftrightarrow\left(x+10\right).\dfrac{1}{9}+\left(x+10\right).\dfrac{1}{8}-\left(x+10\right).\dfrac{1}{7}-\left(x+10\right).\dfrac{1}{6}=0\\ \Leftrightarrow\left(x+10\right)\left(\dfrac{1}{9}+\dfrac{1}{8}-\dfrac{1}{7}-\dfrac{1}{6}\right)=0\)

vì \(\dfrac{1}{9}+\dfrac{1}{8}-\dfrac{1}{7}-\dfrac{1}{6}\ne0\)

nên \(x+10=0\Rightarrow x=-10\)

vậy phương trình có tập nghiệm là S={-10}

Ta có:

\(\dfrac{x+1}{9}+\dfrac{x+2}{8}=\dfrac{x+3}{7}+\dfrac{x+4}{6}\)

\(\Leftrightarrow\dfrac{x+1}{9}+1+\dfrac{x+2}{8}+1=\dfrac{x+3}{7}+1+\dfrac{x+4}{6}+1\)

\(\Leftrightarrow\dfrac{x+1}{9}+\dfrac{9}{9}+\dfrac{x+2}{8}+\dfrac{8}{8}=\dfrac{x+3}{7}+\dfrac{7}{7}+\dfrac{x+4}{6}+\dfrac{6}{6}\)

\(\Leftrightarrow\dfrac{x+1+9}{9}+\dfrac{x+2+8}{8}=\dfrac{x+3+7}{7}+\dfrac{x+4+6}{6}\)

\(\Leftrightarrow\dfrac{x+10}{9}+\dfrac{x+10}{8}=\dfrac{x+10}{7}+\dfrac{x+10}{6}\)

\(\Leftrightarrow\dfrac{x+10}{9}+\dfrac{x+10}{8}-\dfrac{x+10}{7}-\dfrac{x+10}{6}=0\)

\(\Leftrightarrow\left(x+10\right)\left(\dfrac{1}{9}+\dfrac{1}{8}-\dfrac{1}{7}-\dfrac{1}{6}\right)=0\)

\(\Leftrightarrow\left(x+10\right)\cdot-\dfrac{37}{504}=0\)

\(\Rightarrow x+10=0\)

\(\Leftrightarrow x=-10\)

Vậy tập nghiệm của phương trình trên là S={-10}