\(7^x+7^{x+1}+7^{x+2}=3\cdot19\cdot343\)

\(\Leftrightarrow7^x=343\)

hay x=3

I don't now 

sorry 

...................

nha

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

\(\Leftrightarrow\)\(\left[\left(3x-1\right)+\left(2x-1\right)\right]^2=0\)

\(\Leftrightarrow\)\(\left(5x-2\right)^2=0\)

\(\Leftrightarrow\)\(5x-2=0\)

\(\Leftrightarrow\)\(x=\frac{2}{5}\)

Vậy...

b)  \(\left(7x+2\right)^2+\left(7x-2\right)^2-2\left(7x+2\right)\left(7x-2\right)=0\)

\(\Leftrightarrow\)\(\left[\left(7x+2\right)-\left(7x-2\right)\right]^2=0\)

\(\Leftrightarrow\)\(4^2=0\)  vô lí

Vậy pt vô nghiệm

-7X-3X

j

\(\left(7x+3\right)^2-\left(7x-1\right)\left(7x-3\right)=-12\)

\(\Rightarrow49x^2+42x+9-\left(49x^2-21x-7x+3\right)=-12\)

\(\Rightarrow70x+18=0\) \(\Rightarrow x=-\dfrac{18}{70}=-\dfrac{9}{35}\)

do các phân số ở hàng số thứ 2  đã tối giản nên x=0=>7x=0 =>tổng các phân số sau đều tối giản

`@` `\text {Ans}`

`\downarrow`

`7x*(1/7x - 2x^2 + 1)`

`= 7x*1/7x + 7x*(-2x^2) + 7x`

`= x^2 - 14x^3 + 7x`

`= -14x^3 + x^2 + 7x`

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

<=> \(9x^2-9x+2=9x^2+6x+1\)

<=>  \(15x=1\) <=> \(x=\frac{1}{15}\)

b) \(\left(4x-1\right)\left(x+1\right)=\left(2x-3\right)^2\)

<=> \(4x^2+3x-1=4x^2-12x+9\)

<=> \(15x^2=10\) <=> \(x=\frac{2}{3}\)

c) \(\left(5x+1\right)^2=\left(7x-3\right)\left(7x+2\right)\) <=> \(25x^2+10x+1=49x^2-7x-6\)

<=> \(24x^2-17x-7=0\) <=> \(24x^2-24x+7x-7=0\)

<=> \(\left(24x+7\right)\left(x-1\right)=0\) <=> \(\orbr{\begin{cases}x=-\frac{7}{24}\\x=1\end{cases}}\)

d) (4 - 3x)(4 + 3x) = (9x - 3)(1 - x)

<=> 16 - 9x² = 12x - 9x2 - 3

<=> 12x = 19

<=> x = 19/12

e) x(x + 1)(x + 2)(x + 3) = 24

<=> (x² + 3x)(x² + 3x + 2) = 24

<=> (x² + 3x)²  + 2(x² + 3x) - 24 = 0

<=> (x² + 3x)² + 6(x² + 3x) - 4(x² + 3x) - 24 = 0

<=> (x² + 3x + 6)(x² + 3x - 4) = 0

<=> \(\orbr{\begin{cases}x^2+3x+6=0\\x^2+3x-4=0\end{cases}}\)

<=> \(\orbr{\begin{cases}\left(x+\frac{3}{2}\right)^2+\frac{15}{4}=0\left(vn\right)\\\left(x+4\right)\left(x-1\right)=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-4\\x=1\end{cases}}\)

g) (7x - 2)² = (7x - 3)(7x + 2)

<=> 49x² - 28x + 4 = 49x² - 7x - 6

<=> 21x = 10 <=> x = 10/21

Ta có: A = (4x2 - 7x + 1) - (3x2 - 7x - 1) = x2 + 2. Chọn C

Đặt \(\hept{\begin{cases}\sqrt{7x+11}=a\\\sqrt{9-7x}=b\end{cases}}\)

\(\Rightarrow a^2-b^2=14x+2\)

\(\Rightarrow\frac{2}{a^2-b^2}+\frac{1}{ab}=\frac{7}{24}\)

\(\Leftrightarrow\left(b+7a\right)\left(7b-a\right)=0\)

Làm nhầm phần phân tích nhân tử giờ làm lại cách khác.

Đặt \(7x+11=a\)

\(\Rightarrow7x=a-11\)

\(\Rightarrow\frac{1}{a-10}+\frac{1}{\sqrt{a\left(20-a\right)}}=\frac{7}{24}\)

\(\Leftrightarrow\frac{1}{\sqrt{a\left(20-a\right)}}=\frac{7}{24}-\frac{1}{a-10}\)

\(\Leftrightarrow\frac{1}{a\left(20-a\right)}=\left(\frac{7}{24}-\frac{1}{a-10}\right)^2\)

\(\Leftrightarrow\left(a-18\right)\left(a-16\right)\left(49a^2-630a+200\right)=0\)

PS: Bài giải trên bỏ đi nha