chứng minh
\(\sqrt{x^2+\left(x+1\right)^2+\left(x+2\right)^2+...+\left(x+n\right)^2}=x+\left(x+1\right)+\left(x+2\right)+...+\left(x+n\right)\)
Rút gọn:
\(M=1-\left[\dfrac{2x-1+\sqrt{x}}{1-x}+\dfrac{2x\sqrt{x}+x-\sqrt{x}}{1+x\sqrt{x}}\right]\cdot\left[\dfrac{\left(x-\sqrt{x}\right)\left(1-\sqrt{x}\right)}{2\sqrt{x}-1}\right]\)
Giải::
ĐK: x khác +- 1
\(M=1-\left[\dfrac{\left(\sqrt{x}-\dfrac{1}{2}\right)\left(\sqrt{x}+1\right)}{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-\dfrac{1}{2}\right)\left(\sqrt{x}+1\right)}{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}+x\right)}\right]\cdot\left[\dfrac{-\sqrt{x}\left(1-\sqrt{x}\right)^2}{2\left(\sqrt{x}-\dfrac{1}{2}\right)}\right]\)
\(=1-\left[\dfrac{\left(\sqrt{x}-\dfrac{1}{2}\right)}{\left(1-\sqrt{x}\right)}\cdot\dfrac{-\sqrt{x}\left(1-\sqrt{x}\right)^2}{2\left(\sqrt{x}-\dfrac{1}{2}\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-\dfrac{1}{2}\right)}{1-\sqrt{x}+x}\cdot\dfrac{-\sqrt{x}\left(1-\sqrt{x}\right)^2}{2\left(\sqrt{x}-\dfrac{1}{2}\right)}\right]\)
\(=1-\left[\dfrac{-\sqrt{x}\left(1-\sqrt{x}\right)}{2}+\dfrac{-x\left(1-\sqrt{x}\right)^2}{2\left(1-\sqrt{x}+x\right)}\right]\)
rồi làm sao nữa ak?? Tớ có quy đồng lên, tính sơ sơ rồi nhưng thấy kq không gọn.
Câu b là : tìm các số nguyên x để M cũng là số nguyên . Nên tớ nghĩ kq sẽ gọn.
NHỜ MẤY CAO NHÂN RA TAY GIÚP VỚI NHAK ^^!
\(P=\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\frac{\left(1-x\right)^2}{2}\)
\(P=\left(\frac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right)\frac{\left(1-x\right)^2}{2}\)
\(P=\left(\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(x-1\right)\left(\sqrt{x}+1\right)}\right)\frac{\left(x-1\right)^2}{2}\)
\(P=\left(\frac{\left(x-\sqrt{x}-2\right)-\left(x+\sqrt{x}-2\right)}{\left(x-1\right)\left(\sqrt{x}+1\right)}\right)\frac{\left(x-1\right)^2}{2}\)
\(P=\frac{2\sqrt{x}}{\left(x-1\right)\left(\sqrt{x}+1\right)}\frac{\left(x-1\right)^2}{2}\)
\(P=\frac{\sqrt{x}\left(x-1\right)}{\sqrt{x}+1}=\sqrt{x}\left(\sqrt{x}-1\right)=x-\sqrt{x}\)
Chứng minh rằng:\(x^{\left(2^{y+1}\right)}+x^{\left(2^y\right)}+1=\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x^4-x^2+1\right)...\left(x^{\left(2^{y-1}\right)}+x^{\left(2^{y-2}\right)}+1\right)\left(x^{\left(2^y\right)}+x^{\left(2^{y-1}\right)}+1\right)\)với mọi \(x\in N;x>0\)và \(y\in N;y>1\)
Cho x , y , z > 0 thỏa mãn \(\left\{{}\begin{matrix}x^2+y^2+z^2=2\\x+y+z=2\end{matrix}\right.\)
Chứng minh \(P=x\sqrt{\dfrac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}+y\sqrt{\dfrac{\left(1+z^2\right)\left(1+x^2\right)}{x+y^2}+z\sqrt{\dfrac{\left(1+y^2\right)\left(1+x^2\right)}{1+z^2}}}\) không phụ thuộc vào biến .
\(=\frac{x+1}{2\left(x-1\right)}+\frac{2}{2\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\)
=\(\frac{\left(x+1\right).\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{2\sqrt{x}\left(\sqrt{x}-1\right)}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{2\sqrt{x}\left(\sqrt{x}+1\right)}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\)
=\(\frac{x\sqrt{x}+\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{2x-2\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{2x+2\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\)
=\(\frac{x\sqrt{x}+4x+\sqrt{x}}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
=\(\frac{\sqrt{x}\left(x+4\sqrt{x}+1\right)}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
=\(\frac{\sqrt{x}\left(\sqrt{x}+1\right)^2}{2\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
=\(\frac{\sqrt{x}+1}{2\left(\sqrt{x}-1\right)}\)
LƯU Ý: CAP NÀY CHỈ LÀ CAP NHÁP
Chứng minh biểu thức sau ko phụ thuộc vào x:
\(A=x\left(x^2+x+1\right)-x^2\left(x+1\right)-x+5\)
\(B=x\left(2x+1\right)-x^2\left(x+2\right)+x^3-x+3\)
\(C=4\left(6-x\right)+x^2\left(2+3x\right)-x\left(5x-4\right)+3x^2\left(1-x\right)\)
\(D=5\left(3x^{n+1}-y^{n-1}\right)+3\left(x^{n+1}+5y^{n-1}\right)-5\left(3x^{n+1}+2y^{n-1}\right)\)
A=5; B=3; C=24 không phụ thuộc x; câu D thì mong bạn xem lại đề
Chứng minh biểu thức sau ko phụ phuộc vào x
A = \(x\left(x^2+x+1\right)-x^2\left(x+1\right)-x+5\)
B = \(x\left(2x+1\right)-x^2\left(x+2\right)+x^3-x+3\)
C = \(4\left(6-x\right)+x^2\left(2+3x\right)-x\left(5x-4\right)+3x^2\left(1-x\right)\)
D = \(5\left(3x^{n+1}-y^{n-1}\right)+3\left(x^{n+1}+5y^{n-1}\right)-5\left(3x^{n+1}+2y^{n-—}\right)\)
\(A=\left(x^3+x^2+x\right)-\left(x^3+x^2\right)-x+5\)5
\(A=x^3+x^2+x-x^3-x^2-x+5\)
=> A=5
=> A luôn = 5 với mọi x => A không phụ thuộc vào x
\(B=x\left(2x+1\right)-x^2\left(x+2\right)+x^3-x+3\)
\(B=\left(2x^2+x\right)-\left(x^3+2x^2\right)+x^3-x+3\)
\(B=2x^2+x-x^3-2x^2+x^3-x+3\)
=> B= 3
=> B luôn =3 với mọi x => B không phụ thuộc vào x
\(C=4\left(6-x\right)+x^2\left(2+3x\right)-x\left(5x-4\right)+3x^2\left(1-x\right)\)
\(C=24-4x+2x^2+3x^3-5x^2+4x+3x^2-3x^3\)
C=24
=> C=24 với mọi x => C không phụ thuộc vào x
Câu D kí tự cuối có vẻ bạn gõ sai nên mình không làm được, sorry nhiều
A = x(x2 + x + 1) - x2(x + 1) - x + 5
A = x.x2 + x.x + x.1 + (-x2).x + (-x2).1 - x + 5
A = x3 + x2 + x - x3 - x2 - x + 5
A = (x3 - x3) + (x2 - x2) + (x - x) + 5
A = 0 + 0 + 0 + 5
A = 5
Vậy: Biểu thức không phụ thuộc giá trị của biến.
B = x(2x + 1) - x2(x + 2) + x3 - x + 3
B = x.2x + x.1 + (-x2).x + (-x2).2 + x3 - x + 3
B = 2x2 + x - x3 - 2x2 + x3 - x + 3
B = (2x2 - 2x2) + (x - x) + (-x3 + x3) + 3
B = 0 + 0 + 0 + 3
B = 3
Vậy: Biểu thức không phụ thuộc giá trị của biến.
C = 4(6 - x) + x2(2 + 3x) - x(5x - 4) + 3x2(1 - x)
C = 4.6 + 4.(-x) + x2.2 + x2.3x + (-x).5x + (-x).(-4) + 3x2.1 + 3x2.(-x)
C = 24 - 4x + 2x2 + 3x3 - 5x2 + 4x + 3x2 - 3x3
C = 24 + (-4x + 4x) + (2x2 - 5x2 + 3x2) + (3x3 - 3x3)
C = 24 + 0 + 0 + 0
C = 24
Vậy: Biểu thức không phụ thuộc giá trị của biến.
D viết sai thì chịu
\(B=\left(\frac{x\sqrt{x}+x+\sqrt{x}}{x\sqrt{x}-1}-\frac{\sqrt{x}+3}{1-\sqrt{x}}\right).\frac{x-1}{2x+\sqrt{x}-1}\) ĐKXĐ: ...
\(=\frac{\left(x\sqrt{x}+x+\sqrt{x}\right)\left(1-\sqrt{x}\right)-\left(\sqrt{x}+3\right)\left(x\sqrt{x}-1\right)}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{x-1}{2x+2\sqrt{x}-\sqrt{x}-1}\)
\(=\frac{x\sqrt{x}+x+\sqrt{x}-x^2-x\sqrt{x}-x-x^2+\sqrt{x}-3x\sqrt{x}+3}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{x-1}{2\sqrt{x}\left(\sqrt{x}+1\right)-\left(\sqrt{x}+1\right)}\)
\(=\frac{-3x\sqrt{x}+2\sqrt{x}-2x^2+3}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{x-1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{3-3x\sqrt{x}+2\sqrt{x}-2x^2}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{3\left(1-x\sqrt{x}\right)+2\sqrt{x}\left(1-x\sqrt{x}\right)}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(2\sqrt{x}+3\right)\left(1-x\sqrt{x}\right)}{\left(x\sqrt{x}-1\right)\left(1-\sqrt{x}\right)}.\frac{x-1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{-2\sqrt{x}-3}{1-\sqrt{x}}.\frac{x-1}{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{-2\sqrt{x}-3}{1-\sqrt{x}}.\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{2\sqrt{x}-1}\)
\(=\frac{2\sqrt{x}+3}{2\sqrt{x}-1}\)
\(\left(\frac{1}{2+2\sqrt{x}}+\frac{1}{2-2\sqrt{x}}-\frac{x^2+1}{1-x^2}\right)\left(1+\frac{1}{x}\right)\)
\(=\left(\frac{2-2\sqrt{x}+2+2\sqrt{x}}{\left(2+2\sqrt{x}\right)\left(2-2\sqrt{x}\right)}-\frac{x^2+1}{1-x^2}\right)\left(1+\frac{1}{x}\right)\)
\(=\left(\frac{4}{4-4x}-\frac{x^2+1}{\left(1-x\right)\left(1+x\right)}\right)\left(1+\frac{1}{x}\right)\)
\(=\left(\frac{1+x-x^2-1}{\left(1-x\right)\left(1+x\right)}\right)\left(1+\frac{1}{x}\right)=\frac{x\left(1-x\right)}{\left(1-x\right)\left(1+x\right)}.\frac{x+1}{x}=1\)
Nếu bạn bảo kiểm tra thì lời giải đúng rồi nhé!