Chủ đề này có 3 trả lời

[trang2004]

Giai phuong trinh

1, $(\sqrt{x+3}-\sqrt{x-1})(x^2+\sqrt{x^2+4x+3})=2x$

2, $2\sqrt{3x+4}+3\sqrt{5x+9}=x^2+6x+13$

3, $(x+1)\sqrt{x+8}=x^2+x+4$

4, $\sqrt{2x^2+16x+18}+\sqrt{x^2-1}=2x+4$

5, $\sqrt{3-x}+\sqrt{2+x}=x^3+x^2-4x-4+|x|+|x-1|$

6, $\sqrt{\frac{1-x}{x}}=\frac{2x+x^2}{1+x^2}$

[Frosty Flame]

3, $(x+1)\sqrt{x+8}=x^2+x+4$

ĐKXĐ: $x\geq -8$

Vì $(x+1)\sqrt{x+8}=x^2+x+4>0=>x+1>0<=>x>-1$

Pt đã cho

$<=>(x+1)^2(x+8)=(x^2+x+4)^2<=>...<=>(x-1)(x^3+2x^2+x-8)=0$

$<=>x=1 hoặc x^3+2x^2+x-8=0$

Xét $x^3+2x^2+x-8=0$

Đặt $x=t-\frac{3}{2}(t-\frac{3}{2}>1<=>t> \frac{-1}{3})$

Thay vào

$=>...<=>27t^3-9t=218$

Đặt $3t=y(t>\frac{-1}{3}<=>y>-1)$

$=>y^3-3y=218

$+)$Xét $-1y<2=>y^3-3y=(y-2)(y+1)^2+2<2<218 ( Loại )$

$+)$Xét $y\geq 2$

Lúc này, ta có thể đặt $y=a+\frac{1}{a}(a>0)$

Thay vào

$=>(a+\frac{1}{a})^3-3(a+\frac{1}{a})=0<=>a^3+\frac{1}{a^3}=218$

$<=>a^6-218a^3+1=0$

$<=>...<=>a^3=109\pm 6\sqrt{330}$

$<=>a=\sqrt[3]{109\pm 6\sqrt{330}}$

$<=>...<=>x=\frac{\sqrt[3]{109-6\sqrt{330}}+\sqrt[3]{109+6\sqrt{330}}-2}{3}$

Thử lại thấy thỏa mãn.

Vậy$,...$

[Frosty Flame]

Giai phuong trinh

1, $(\sqrt{x+3}-\sqrt{x-1})(x^2+\sqrt{x^2+4x+3})=2x$

ĐKXĐ: $x\geq -1$

Pt đã cho

$<=>\frac{2}{\sqrt{x+3}+\sqrt{x+1}}[x^2+\sqrt{(x+1)(x+3)}]=2x$

$<=>x^2+\sqrt{(x+1)(x+3)}=x\sqrt{x+1}+x\sqrt{x+3}<=>(x-\sqrt{x+1})(x-\sqrt{x+3})=0<=>...$

 

Giai phuong trinh

2, $2\sqrt{3x+4}+3\sqrt{5x+9}=x^2+6x+13$

ĐKXĐ: $x\geq \frac{-4}{3}$

Pt đã cho

$<=>x^2+x+2[(x+2)-\sqrt{3x+4}]+3[(x+3)-\sqrt{5x+9}]=0$

$<=>x(x+1)[1+\frac{2}{x+2+\sqrt{3x+4}}+\frac{3}{x+3+\sqrt{5x+9}}]=0<=>...$

 

4, $\sqrt{2x^2+16x+18}+\sqrt{x^2-1}=2x+4$

ĐKXĐ: ...

Pt đã cho

$<=>2x^2+16x+18+x^2-1+2\sqrt{(2x^2+16x+18)(x^2-1)}=(2x+4)^2$

$<=>x^2-1-2\sqrt{(2x^2+16x+18)(x^2-1)}=0<=>\sqrt{x^2-1}(\sqrt{x^2-1}-2\sqrt{2x^2+16x+18})=0<=>...$

[Frosty Flame]

5, $\sqrt{3-x}+\sqrt{2+x}=x^3+x^2-4x-4+|x|+|x-1|$

ĐKXĐ: $-2\leq x\leq 3$

Pt đã cho

$<=>(x+2)(x^2-x-2)+\begin{vmatrix}x\end{vmatrix}-\sqrt{2+x}+\begin{vmatrix}x-1\end{vmatrix}-\sqrt{3-x}=0$

$<=>...<=>x^2-x-2=0<=>...$

 

6, $\sqrt{\frac{1-x}{x}}=\frac{2x+x^2}{1+x^2}$

ĐKXĐ: $0<x\leq 1$

Pt đã cho

$<=>(1-x)(1+x^2)^2=x(2x+x^2)^2<=>...<=>(2x-1)(x^4+2x^3+4x^2+x+1)=0<=>...$