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![](https://rs.olm.vn/images/avt/0.png?1311)
1) \(\Rightarrow16x^2+24x+9+9x^2-24x+16+4-25x^2=x\)
\(\Rightarrow x=29\)
2)
a) \(=x^2-9-x^2+6x-9=6x-18\)
b) \(=\left(3x-1+2x+1\right)^2=\left(5x\right)^2=25x^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Mk xin phép ko vt lại đề nx
\(\Rightarrow A=\left[\left(3x-2\right)\left(x+1\right)-\left(2x+5\right)\left(x^2-1\right)\right]\div x+1\)
\(\Rightarrow A=3x-2-\left(2x-5\right)\left(x-1\right)\)
\(\Rightarrow x=\dfrac{1}{2}\)
\(\Rightarrow A=\dfrac{3}{2}-2-\left(1-5\right)\left(\dfrac{1}{2}-1\right)=-\dfrac{5}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,A=\left(x+y\right)^2-9z^2=\left(x+y-3z\right)\left(x+y+3z\right)\\ A=\left(5+7-36\right)\left(5+7+36\right)=-24\cdot48=-1152\\ b,B=\left(2x-y\right)\left(2x+y\right)+\left(2x+y\right)=\left(2x+y\right)\left(2x-y-1\right)\\ B=\left(2+2\right)\left(2-2-1\right)=4\cdot\left(-1\right)=-4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`a) x(x + 5)(x – 5) – (x + 2)(x^2 – 2x + 4) = 3`
`<=>x(x^2-25)-(x^3-8)=3`
`<=>x^3-25x-x^3+8=3`
`<=>-25x=-5`
`<=>x=1/5`
`b) (x – 3)^3 – (x – 3)(x^2 + 3x + 9) + 9(x + 1)^2 = 15`
`<=>x^3-9x^2+27x-27-(x^3-27)+9(x^2+2x+1)=15`
`<=>-9x^2+27x+9x^2+18x+9=15`
`<=>45x+9=15`
`<=>45x=6`
`<=>x=6/45=2/15`
`c) (x+5)(x^2 –5x +25) – (x – 7) = x^3`
`<=>x^3-125-x+7=x^3`
`<=>x^3-x-118=x^3`
`<=>-x-118=0`
`<=>-x=118<=>x=-118`
`d) (x+2)(x^2 – 2x + 4) – x(x^2 + 2) = 4 `
`<=>x^3+8-x^3-2x=4`
`<=>8-2x=4`
`<=>2x=4<=>x=2`
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Điều kiện: x + 2 ≠ 0 và x – 2 ≠ 0 ⇔ x ≠ ± 2
(Khi đó: x2 – 4 = (x + 2)(x – 2) ≠ 0)
Vậy tập nghiệm của pt là: S = {-1; 1}
b) Điều kiện: 2x ≥ 0 ⇔ x ≥ 0
Khi đó: |x – 5| = 2x ⇔ x – 5 = 2x hoặc x – 5 = -2x
⇔ x = -5 hoặc x = 5/3
Vì x ≥ 0 nên ta lấy x = 5/3 . Tập nghiệm : S = {5/3}
c) x – 2)2 + 2(x – 1) ≤ x2 + 4
⇔ x2 – 4x + 4 + 2x – 2 ≤ x2 + 4
⇔ -2x ≤ 2
⇔ x ≥ -1
Tập nghiệm S = {x | x ≥ -1}
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu a,b hình như nhầm đề mình tự sửa nha ;-;
a, Ta có : \(\left(x^2-x-6\right)^2+\left(x-3\right)^2\)
\(=\left(x^2-3x+2x-6\right)^2+\left(x-3\right)^2\)
\(=\left(x-3\right)^2\left(x+2\right)^2+\left(x-3\right)^2\)
\(=\left(x-3\right)^2\left(\left(x+2\right)^2+1\right)\)
b, Ta có : \(\left(x^2-x-20\right)^2+\left(x+4\right)^2\)
\(=\left(x^2+4x-5x-20\right)^2+\left(x+4\right)^2\)
\(=\left(x+4\right)^2\left(x-5\right)^2+\left(x+4\right)^2\)
\(=\left(x+4\right)^2\left(\left(x-5\right)^2+1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
a: Ta có: \(A=\left(x+1\right)^3+\left(x-1\right)^3\)
\(=x^3+3x^2+3x+1+x^3-3x^2+3x-1\)
\(=2x^3+6x\)
b: Ta có: \(B=\left(x-3\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+\left(3x-1\right)\left(3x+1\right)\)
\(=x^3-9x^2+27x-27-x^3-27+9x^2-1\)
\(=27x-55\)
Vì \(\sqrt{x}\ge0\Rightarrow0\le\sqrt{x}< 3\)
\(\Rightarrow0\le x< 9\Rightarrow x\in\left\{0;1;2;...;8\right\}\)
\(b,x^2=5\)
\(\Rightarrow\sqrt{x^2}=\pm\sqrt{5}\)
\(\Rightarrow x=\pm\sqrt{5}\)
c, Ta có:\(x\ne0\left(\sqrt{x}\ge0\forall x\right)\)
\(\Rightarrow\sqrt{0}\le\sqrt{x}< \sqrt{2}\)
\(\Rightarrow0\le x< 4\)
\(\Rightarrow x\in\left\{0;1;2;3\right\}\)