a. 6x² - (2x + 5)(3x - 2) = 7

<=> 6x² - 6x² + 4x - 15x + 10 = 7

<=> -11x = -3

<=> \(x=\dfrac{3}{11}\)

b. (5 - x)(25 + 5x + x²) + x(x² - 7) = 25

<=> 125 - x³ + x³ - 7x = 25

<=> -7x = 25 - 125

<=> -7x = -100

<=> \(x=\dfrac{100}{7}\)

c. (7 - 2x)² + (3 + 2x)(3 - 2x) = 30

<=> 49 - 28x + 4x² + 9 - 4x² = 30

<=> 4x² - 4x² - 28x = 30 - 49 - 9

<=> -28x = -28

<=> x = 1

Bài 3:

a:

ĐKXĐ: x>=-5

\(x^2-7x=6\sqrt{x+5}-30\)

=>\(x^2-4x-3x+12=6\sqrt{x+5}-18\)

=>\(\left(x-4\right)\left(x-3\right)=6\left(\sqrt{x+5}-3\right)\)

=>\(\left(x-4\right)\left(x-3\right)=6\cdot\frac{x+5-9}{\sqrt{x+5}+3}\)

=>\(\left(x-4\right)\left(x-3-\frac{6}{\sqrt{x+5}+3}\right)=0\)

=>x-4=0

=>x=4(nhận)

Bài 2:

a: ĐKXĐ: x>=0

\(\sqrt{x+4\sqrt{x}+4}=5x+2\)

=>\(\sqrt{\left(\sqrt{x}+2\right)^2}=5x+2\)

=>\(5x+2=\sqrt{x}+2\)

=>\(5x-\sqrt{x}=0\)

=>\(\sqrt{x}\left(5\sqrt{x}-1\right)=0\)

=>\(\left[\begin{array}{l}\sqrt{x}=0\\ 5\sqrt{x}-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ \sqrt{x}=\frac15\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\left(nhận\right)\\ x=\frac{1}{25}\left(nhận\right)\end{array}\right.\)

b: \(\sqrt{x^2-2x+1}+\sqrt{x^2+4x+4}=4\)

=>\(\sqrt{\left(x-1\right)^2}+\sqrt{\left(x+2\right)^2}\) =4

=>|x-1|+|x+2|=4(1)

TH1: x<-2

(1) sẽ trở thành: -x-2+1-x=4

=>-2x-1=4

=>-2x=5

=>x=-5/2(nhận)

TH2: -2<=x<1

(1) sẽ trở thành: x+2+1-x=4

=>3=4(vô lý)

TH3: x>=1

(1) sẽ trở thành: x+2+x-1=4

=>2x+1=4

=>2x=3

=>x=3/2(nhận)

c: ĐKXĐ: x>=1

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

=>\(\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}=2\)

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

=>\(\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|=2\)

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

=>\(\sqrt{x-1}-1\le0\)

=>\(\sqrt{x-1}\le1\)

=>0<=x-1<=1

=>1<=x<=2

a)\(\left(x2+7\right).\left(x2-49\right)< 0\)

\(\left(x2+7\right).\left(x2-49\right)< 0\) chứng tỏ hai vế \(\left(x2+7\right)\) và \(\left(x2-49\right)\) khác dấu nhau .

\(\left\{{}\begin{matrix}\left(x2+7\right)>0\\\left(x2-49\right)< 0\end{matrix}\right.\)

Vì \(\left(x2+7\right)\) > \(\left(x2-49\right)\)

Nên ta có:

\(\left\{{}\begin{matrix}\left(x2+7\right)>0\\\left(x2-49\right)< 0\end{matrix}\right.\)\(\Rightarrow\)\(\left\{{}\begin{matrix}\left(x+7\right)=0\\\left(x-49\right)=0\end{matrix}\right.\)\(\Rightarrow\)\(\left\{{}\begin{matrix}x=-7\\x=49\end{matrix}\right.\)

Vậy hai số nguyên đó là -7 và 49 .

Còn phần còn lại bạn làm tương tự nhé !

Bài 1)1)\(x^2+5x+6=x^2+3x+2x+6\)=0

=x(x+3)+2(x+3)=(x+2)(x+3)=0

Dễ rồi

2)\(x^2-x-6=0=x^2-3x+2x-6=0\)

=x(x-3)+2(x-3)=0

=(x+2)(x-3)=0

Dễ rồi

3)Phương trình tương đương:\(\left(x^2+1\right)\left(x+2\right)^2=0\)

Vì \(x^2+1>0\)

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

Dễ rồi

4)Phương trình tương đương\(x^2\left(x+1\right)+\left(x+1\right)\)=0

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

=>x+1=0

=>..................

5)\(x^2-7x+6=x^2-6x-x+6\) =0

=x(x-6)-(x-6)=0

=(x-1)(x-6)=0

=>.....

6)\(2x^2-3x-5=2x^2+2x-5x-5\)=0

=2x(x+1)-5(x+1)=0

=(2x-5)(x+1)=0

7)\(x^2-3x+4x-12\)=x(x-3)+4(x-3)=(x+4)(x-3)=0

Dễ rồi

Nghỉ đã hôm sau làm mệt

Ta có \(x1-\frac{1}{9}=x2-\frac{2}{8}=...=x9-\frac{9}{1}\)

\(=\frac{x1-1}{9}=\frac{x2-2}{8}=\frac{x3-3}{7}=...=\frac{x9-9}{1}\)

= \(\frac{x1-1+x2-2+x3-3+...+x9-9}{9+8+7+...+1}\)

\(=\frac{\left(x1+x2+x3+...+x9\right)-\left(1+2+3+...+9\right)}{9+8+7+....+1}\)

=\(\frac{90-45}{45}=\frac{45}{45}=1\)

=> \(\hept{\begin{cases}\begin{cases}x1=10\\x2=10\end{cases}\\.....\\x9=10\end{cases}}\)