a, \(\sqrt{\left(x-1\right)^2}=5\Rightarrow\left(x-1\right)=\left\{5;-5\right\}\Leftrightarrow\hept{\begin{cases}x-1=5\Rightarrow x=6\\x-1=-5\Rightarrow x=-4\end{cases}}\)

b,\(3+\sqrt{x}=5\Rightarrow\sqrt{x}=2\Rightarrow x=4\)

c,\(\sqrt{x^2-2x+1}=x-1\Rightarrow\sqrt{\left(x-1\right)^2}=x-1\Rightarrow x-1=\left\{x-1;-\left(x-1\right)\right\}\)

\(\Leftrightarrow\hept{\begin{cases}x-1=x-1\Rightarrow x\in R\\x-1=-\left(x-1\right)\Rightarrow x-1=-x+1\Rightarrow x+x=1+1\Rightarrow2x=2\Rightarrow x=1\end{cases}}\)

Vậy x = 1

d, \(\sqrt{x^2-10x+25}=x+3\Rightarrow\sqrt{\left(x-5\right)^2}=x+3\Rightarrow x-5=\left\{x+3;-\left(x+3\right)\right\}\)

\(\Leftrightarrow\hept{\begin{cases}x-5=x+3\Rightarrow x-x=3+5\Rightarrow0x=8\left(loai\right)\\x-5=-\left(x+3\right)\Rightarrow x-5=-x-3\Rightarrow x+x=-3+5\Rightarrow2x=2\Rightarrow x=1\left(chon\right)\end{cases}}\)

Vậy x = 1

\(xy+\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=2019\)
\(\Leftrightarrow x^2y^2+\left(1+x^2\right)\left(1+y^2\right)+2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=2019^2\)

\(\Leftrightarrow M=2xy\sqrt{\left(1+x^2\right)\left(1+y^2\right)}=2019^2-2x^2y^2-x^2-y^2-1\)(Đặt M = .... cho gọn)

Có \(S=x\sqrt{1+y^2}+y\sqrt{1+x^2}\)

\(\Rightarrow S^2=x^2\left(1+y^2\right)+y^2\left(1+x^2\right)+M\)

\(\Rightarrow S^2=2x^2y^2+x^2+y^2+2019^2-2x^2y^2-x^2-y^2-1\)

\(\Rightarrow S=\sqrt{2019^2-1}\)

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\(a,\)\(\left(4x-2\right)\left(10x+4\right)\left(5x+7\right)\left(2x+1\right)+17\)

\(=\left(4x-2\right)\left(5x+7\right)\left(10x+4\right)\left(2x+1\right)+17\)

\(=\left(20x^2+18x-5\right)\left(20x^2+18x+4\right)+17\)

Đặt ....

Câu 1:

ĐKXĐ: $3\geq x\geq -2$

PT \(\sqrt{x+2}-2-(\sqrt{3-x}-1)=x^2-6x+8\)

\(\Leftrightarrow \frac{x-2}{\sqrt{x+2}+2}-\frac{2-x}{\sqrt{3-x}+1}=(x-2)(x-4)\) (liên hợp)

\(\Leftrightarrow (x-2)\left[\frac{1}{\sqrt{x+2}+2}+\frac{1}{\sqrt{3-x}+1}-x+4\right]=0\)

Ta thấy với mọi $3\geq x\geq -2$ thì:

\(\frac{1}{\sqrt{x+2}+2}+\frac{1}{\sqrt{3-x}+1}>0\)

\(-x+4>0\)

\(\Rightarrow \frac{1}{\sqrt{x+2}+2}+\frac{1}{\sqrt{3-x}+1}-x+4>0\)

\(\Rightarrow \frac{1}{\sqrt{x+2}+2}+\frac{1}{\sqrt{3-x}+1}-x+4\neq 0\)

Do đó $x-2=0$ hay PT có nghiệm duy nhất $x=2$ (t/m)

Bài 3:

ĐK: $1\geq x\geq -1$.

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

\(\Leftrightarrow x\left(\frac{\sqrt{1-x}+1}{\sqrt{1+x}+1}-2\right)=0\)

\(\Rightarrow \left[\begin{matrix} x=0(\text{thỏa mãn})\\ \frac{\sqrt{1-x}+1}{\sqrt{1+x}+1}=2(1)\end{matrix}\right.\)\(\Rightarrow \left[\begin{matrix} x=0(\text{thỏa mãn})\\ \sqrt{1-x}+1=2\sqrt{1+x}+2(1)\end{matrix}\right.\)

Câu 6:

ĐK: \(x\leq 0\) hoặc $x\geq 23$

PT \(\Rightarrow 2x^2-9x+2\sqrt{(x^2-2x)(x^2-7x)}=x^2-23x\) (bình phương 2 vế)

\(\Leftrightarrow x^2+14x+2\sqrt{x^2(x-2)(x-7)}=0(*)\)

Nếu $x\geq 23$ thì hiển nhiên vế trái của $(*)$ dương (vô lý). Do đó $x\leq 0$

Khi đó \((*)\Leftrightarrow x^2+14x-2x\sqrt{(x-2)(x-7)}=0\)

\(\Leftrightarrow x(x+14-2\sqrt{(x-2)(x-7)})=0\)

\(\Rightarrow \left[\begin{matrix} x=0(\text{thỏa mãn})\\ x+14=2\sqrt{(x-2)(x-7)}(2)\end{matrix}\right.\)

Với $(2)$: \(\Rightarrow \left\{\begin{matrix} x+14\geq 0\\ (x+14)^2=4(x-2)(x-7)\end{matrix}\right.\)

\(\Rightarrow \left\{\begin{matrix} x\geq -14\\ 3x^2-64x-140=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=-2\\ x=\frac{70}{3}\end{matrix}\right.\)

Kết hợp với $x\leq 0$ suy ra $x=-2$

Vậy $x=-2$ hoặc $x=0$

Dat \(a=\sqrt[3]{65+x},b=\sqrt[3]{65-x}\)

Bien doi PT thanh \(a^2+4b^2=5ab\)

\(\Leftrightarrow a^2-5ab+4b^2=0\)

\(\Leftrightarrow\left(a^2-ab\right)-\left(4ab-4b^2\right)=0\)

\(\Leftrightarrow a\left(a-b\right)-4b\left(a-b\right)=0\)

\(\Leftrightarrow\left(a-b\right)\left(a-4b\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=b\left(1\right)\\a=4b\left(2\right)\end{cases}}\)

\(\left(1\right)\Leftrightarrow\sqrt[3]{65+x}=\sqrt[3]{65-x}\)

\(\Leftrightarrow65+x=65-x\)

\(\Leftrightarrow x=0\left(n\right)\)

\(\left(2\right)\Leftrightarrow\sqrt[3]{65+x}=4\sqrt[3]{65-x}\)

\(\Leftrightarrow65+x=64.65-64x\)

\(\Leftrightarrow65x=64.65-65\)

\(\Leftrightarrow x=63\left(n\right)\)

Vay nghiem cua PT la \(x=0,x=63\)

Bài 1: 

a ) = 12/21

b ) = 50

k cho mik nha

Các bn giải cụ thể ra giúp mk đc k? c. ơn các bn